SlideShare a Scribd company logo
1 of 8
Download to read offline
Theorems
Midpoint Theorem 𝐼𝑓 𝐵 𝑖𝑠 𝑡ℎ𝑒 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐶
̅̅̅̅, 𝑡ℎ𝑒𝑛 𝐴𝐵
̅̅̅̅ ≅ 𝐵𝐶
̅̅̅̅
Segment Congruence Theorems
Reflexive
Symmetric
Transitive
𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑒 𝐴𝑙𝑔𝑒𝑏𝑟𝑎𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠
𝑜𝑓 𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒, 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐, 𝑎𝑛𝑑 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒
𝐴𝐵
̅̅̅̅̅ ≅ 𝐴𝐵
̅̅̅̅
𝐼𝑓 𝐴𝐵
̅̅̅̅ ≅ 𝐶𝐷
̅̅̅̅, 𝑡ℎ𝑒𝑛 𝐶𝐷
̅̅̅̅ ≅ 𝐴𝐵
̅̅̅̅
𝐼𝑓 𝐴𝐵
̅̅̅̅ ≅ 𝐶𝐷
̅̅̅̅ 𝑎𝑛𝑑 𝐶𝐷
̅̅̅̅ ≅ 𝐸𝐹
̅̅̅̅, 𝑡ℎ𝑒𝑛 𝐴𝐵
̅̅̅̅ ≅ 𝐸𝐹
̅̅̅̅
𝐴𝐵
̅̅̅̅ ≅ 𝐶𝐷
̅̅̅̅ 𝑎𝑛𝑑 𝐶𝐷
̅̅̅̅ ≅ 𝐸𝐹
̅̅̅̅, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐴𝐵
̅̅̅̅ ≅ 𝐸𝐹
̅̅̅̅
(𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑖𝑠 𝑢𝑠𝑒𝑑 𝑤ℎ𝑒𝑛 𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑜𝑡ℎ𝑒𝑟)
Supplement Theorem 𝐼𝑓 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑝𝑎𝑖𝑟 𝑖𝑠 𝑓𝑜𝑟𝑚𝑒𝑑 𝑏𝑦 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠, 𝑡ℎ𝑒𝑛
𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦.
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
(𝑠𝑢𝑚 𝑜𝑓 𝑚∠1 𝑎𝑛𝑑 𝑚∠2 𝑒𝑞𝑢𝑎𝑙𝑠 180°)
Complement Theorem 𝐼𝑓 𝑡ℎ𝑒 𝑢𝑛𝑐𝑜𝑚𝑚𝑜𝑛 𝑠𝑖𝑑𝑒𝑠 𝑓𝑟𝑜𝑚 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠
𝑓𝑜𝑟𝑚 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜
𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦.
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
(𝑠𝑢𝑚 𝑜𝑓 𝑚∠1 𝑎𝑛𝑑 𝑚∠2 𝑒𝑞𝑢𝑎𝑙𝑠 90°)
(𝑝𝑢𝑟𝑝𝑙𝑒 𝑏𝑜𝑥 𝑠𝑦𝑚𝑏𝑜𝑙 𝑚𝑒𝑎𝑛𝑠 𝑏𝑙𝑎𝑐𝑘 𝑙𝑖𝑛𝑒𝑠 𝑎𝑟𝑒 𝑎
𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 𝑜𝑟 90°)
Congruent Supplement Theorem
(3 angles)
Congruent Supplement Theorem
(4 angles)
𝐼𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑑 𝑎𝑛𝑔𝑙𝑒 𝑜𝑛𝑒
𝑖𝑠 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑤𝑖𝑡ℎ 𝑎 𝑡ℎ𝑖𝑟𝑑 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛
𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑎𝑛𝑔𝑙𝑒
𝑖𝑠 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑛𝑔𝑙𝑒.
𝐼𝑓 𝑡𝑤𝑜 𝑠𝑒𝑡𝑠 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒𝑠 𝑐𝑟𝑒𝑎𝑡𝑒 𝑡𝑤𝑜 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
𝑝𝑎𝑖𝑟𝑠, 𝑎𝑛𝑑 𝑜𝑛𝑒 𝑎𝑛𝑔𝑙𝑒 𝑓𝑟𝑜𝑚 𝑒𝑎𝑐ℎ 𝑠𝑒𝑡 𝑖𝑠 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑡𝑜
𝑎𝑛 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑒𝑡, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑛𝑜𝑡 𝑙𝑎𝑏𝑒𝑙𝑒𝑑
𝑎𝑠 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑙𝑠𝑜.
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠1 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 ∠2 ≅ ∠3
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠3 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠2 ≅ ∠3
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, ∠1 ≅ ∠4
Congruent Complement Theorem 𝐼𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑑 𝑎𝑛𝑔𝑙𝑒 𝑜𝑛𝑒
𝑖𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑤𝑖𝑡ℎ 𝑎 𝑡ℎ𝑖𝑟𝑑 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛
𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑛𝑔𝑙𝑒.
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, ∠1 ≅ ∠3
Vertical Angles Theorem 𝐼𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑛𝑔𝑙𝑒𝑠,
𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡.
∠1 ≅ ∠3
∠2 ≅ ∠4
Right Angle Theorems
Perpendicular Lines Intersect to
Form Four Right Angles
All Right Angles are Congruent
Perpendicular Lines Will Form
Four Congruent Adjacent Angles
If Two Angles are Both Congruent
and Supplementary, Then Each
Angle Will be a Right Angle
If Two Angles of a Linear Pair are
Congruent, Then They Are Right
Angles
𝐼𝑓 𝑡𝑤𝑜 𝑙𝑖𝑛𝑒𝑠 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟, 𝑡ℎ𝑒𝑛 𝑓𝑜𝑢𝑟 𝑟𝑖𝑔ℎ𝑡
𝑎𝑛𝑔𝑙𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡.
𝐼𝑓 𝑡𝑤𝑜 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠,
𝑡ℎ𝑒𝑛 𝑎𝑙𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡.
𝐼𝑓 𝑡𝑤𝑜 𝑙𝑖𝑛𝑒𝑠 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑡𝑜 𝑎𝑡 𝑎 90° 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛
𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑓𝑜𝑟𝑚𝑒𝑑.
𝐼𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑖𝑛 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑎𝑛𝑑
𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑠 180°, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡
𝑒𝑎𝑐ℎ 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 90°
𝐼𝑓 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑝𝑎𝑖𝑟 𝑖𝑠 𝑓𝑜𝑟𝑚𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡𝑤𝑜 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡
𝑎𝑛𝑔𝑙𝑒𝑠, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑏𝑜𝑡ℎ 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 90°
𝐹𝑜𝑢𝑟 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑟𝑒𝑎𝑡𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒
𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐿𝑖𝑛𝑒𝑠
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑒𝑞𝑢𝑎𝑙 90°
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
∠3 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
∠4 𝑎𝑛𝑑 ∠1 𝑎𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒 𝑃𝑎𝑖𝑟𝑠
𝑎𝑟𝑒 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝑡ℎ𝑒 𝑝𝑎𝑖𝑟𝑠 𝑎𝑟𝑒
𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦, 𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 𝑒𝑎𝑐ℎ 𝑎𝑛𝑔𝑙𝑒
𝑖𝑠 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒.
𝐴𝑙𝑙 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒 𝑃𝑎𝑖𝑟𝑠
𝑎𝑟𝑒 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝑐𝑟𝑒𝑎𝑡𝑒 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟𝑠
𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 𝑒𝑎𝑐ℎ 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒.
Theorem Proofs
Midpoint Theorem: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: 𝐴𝐵
̅̅̅̅ ≅ 𝐵𝐶
̅̅̅̅
𝐵 𝑖𝑠 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐶
̅̅̅̅ 𝑔𝑖𝑣𝑒𝑛
𝐴𝐵 = 𝐵𝐶 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑎 𝑆𝑒𝑔𝑚𝑒𝑛𝑡
𝐴𝐵
̅̅̅̅ ≅ 𝐵𝐶
̅̅̅̅ 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒
Segment Congruence Theorems:
Reflexive Property: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: 𝐴𝐵
̅̅̅̅̅ ≅ 𝐴𝐵
̅̅̅̅
𝐴𝐵
̅̅̅̅ ≅ 𝐴𝐵
̅̅̅̅ 𝑔𝑖𝑣𝑒𝑛
𝐴𝐵 = 𝐴𝐵 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒
𝐴𝐵 = 𝐴𝐵 𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒
𝐴𝐵
̅̅̅̅ ≅ 𝐴𝐵
̅̅̅̅ 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒
Symmetric Property: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: 𝐶𝐷
̅̅̅̅ ≅ 𝐴𝐵
̅̅̅̅
𝐴𝐵
̅̅̅̅ ≅ 𝐶𝐷
̅̅̅̅ 𝑔𝑖𝑣𝑒𝑛
𝐴𝐵 = 𝐶𝐷 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒
𝐶𝐷 = 𝐴𝐵 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝐶𝐷
̅̅̅̅ ≅ 𝐴𝐵
̅̅̅̅ 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒
Transitive Property: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: 𝐴𝐵
̅̅̅̅ ≅ 𝐸𝐹
̅̅̅̅
𝐴𝐵
̅̅̅̅ ≅ 𝐶𝐷
̅̅̅̅ 𝑎𝑛𝑑 𝐶𝐷
̅̅̅̅ ≅ 𝐸𝐹
̅̅̅̅ 𝑔𝑖𝑣𝑒𝑛
𝐴𝐵 = 𝐶𝐷 𝑎𝑛𝑑 𝐶𝐷 = 𝐸𝐹 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒
𝐴𝐵 = 𝐸𝐹
𝐴𝐵 = 𝐸𝐹
𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑜𝑟,
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 (𝑠𝑢𝑏 𝐴𝐵 𝑓𝑜𝑟 𝐶𝐷)
𝐴𝐵
̅̅̅̅ ≅ 𝐸𝐹
̅̅̅̅ 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒
Supplement Theorem: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠1 𝑎𝑛𝑑 ∠2 𝑓𝑜𝑟𝑚 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑝𝑎𝑖𝑟 𝑔𝑖𝑣𝑒𝑛
𝑁𝑜𝑡𝑒: 𝐻𝑜𝑤 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑎 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟?
𝑇𝑤𝑜 𝐴𝑛𝑔𝑙𝑒𝑠 𝑡ℎ𝑎𝑡:
𝐴𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
𝑈𝑛𝑠ℎ𝑎𝑟𝑒𝑑 𝑠𝑖𝑑𝑒𝑠 𝑓𝑜𝑟𝑚 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑟𝑎𝑦𝑠 (180°)
𝑚∠1 + 𝑚∠2 = 180° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠
Complement Theorem: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠
∠1 𝑎𝑛𝑑 ∠2 𝑓𝑜𝑟𝑚 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
𝑔𝑖𝑣𝑒𝑛
𝑁𝑜𝑡𝑒: 𝐻𝑜𝑤 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠?
𝑃𝑎𝑖𝑟 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒𝑠 𝑡ℎ𝑎𝑡 𝑆ℎ𝑎𝑟𝑒 𝑎 𝑉𝑒𝑟𝑡𝑒𝑥,
𝑆ℎ𝑎𝑟𝑒 𝑎 𝑆𝑖𝑑𝑒, 𝑎𝑛𝑑 𝑎𝑟𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑆𝑎𝑚𝑒 𝑃𝑙𝑎𝑛𝑒
𝐻𝑜𝑤 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒?
𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 = 90° 𝑎𝑛𝑔𝑙𝑒
𝑚∠1 + 𝑚∠2 = 90° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠
Alternative Complement Theorem: ∠1 𝑎𝑛𝑑 ∠2 𝑐𝑟𝑒𝑎𝑡𝑒 ∠𝐴𝐵𝐶, 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠𝐴𝐵𝐶 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
∠1 𝑎𝑛𝑑 ∠2 𝑓𝑜𝑟𝑚 ∠𝐴𝐵𝐶
𝑔𝑖𝑣𝑒𝑛
𝑚∠𝐴𝐵𝐶 = 90° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
𝑚∠1 + 𝑚∠2 = 𝑚∠𝐴𝐵𝐶 𝐴𝑛𝑔𝑙𝑒 𝐴𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝑃𝑜𝑠𝑡𝑢𝑙𝑎𝑡𝑒
𝑚∠1 + 𝑚∠2 = 90° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝑠𝑢𝑏 90° 𝑖𝑛 𝑓𝑜𝑟 ∠𝐴𝐵𝐶
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠
Congruent Supplement Theorems:
Congruent Supplement Theorem (3 angles): 𝑃𝑟𝑜𝑣𝑒 ∠2 ≅ ∠3
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠1 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
𝑔𝑖𝑣𝑒𝑛
𝑚∠1 + 𝑚∠2 = 180°
𝑚∠1 + 𝑚∠3 = 180°
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠
180° = 𝑚∠1 + 𝑚∠3 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝑚∠1 + 𝑚∠2 = 𝑚∠1 + 𝑚∠3 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦:
𝐼𝑓 𝑁 = 𝑃 𝑎𝑛𝑑 𝑃 = 𝑀 𝑡ℎ𝑒𝑛, 𝑁 = 𝑀
𝑚∠1 + 𝑚∠2 = 180° = 𝑚∠1 + 𝑚∠3
𝑁 = 𝑃 = 𝑀
𝑚∠2 = 𝑚∠3
𝑆𝑡𝑒𝑝𝑠:
𝑚∠1 + 𝑚∠2 = 𝑚∠1 + 𝑚∠3
𝑚∠1 + 𝑚∠2 − 𝑚∠1 = 𝑚∠1 + 𝑚∠3 − 𝑚∠1
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦:
𝑃 = 𝑚∠1 (𝑃 𝑖𝑠 𝑟𝑒𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑡𝑜 𝑒𝑞𝑢𝑎𝑙 𝑚∠1)
𝑁 = 𝑀
𝑁 − 𝑃 = 𝑀 − 𝑃
∠2 ≅ ∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
Congruent Supplement Theorem (4 angles): 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠2 ≅ ∠3
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑔𝑙𝑒𝑠
∠3 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑔𝑙𝑒𝑠
∠1 ≅ ∠4
𝑔𝑖𝑣𝑒𝑛
𝑚∠1 + 𝑚∠2 = 180°
𝑚∠3 + 𝑚∠4 = 180°
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠
180° = 𝑚∠3 + 𝑚∠4 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝑚∠1 + 𝑚∠2 = 𝑚∠3 + 𝑚∠4 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦:
𝐼𝑓 𝑁 = 𝑃 𝑎𝑛𝑑 𝑃 = 𝑀, 𝑡ℎ𝑒𝑛 𝑁 = 𝑀
𝑚∠1 + 𝑚∠2 = 180° = 𝑚∠3 + 𝑚∠4
𝑁 = 𝑃 = 𝑀
∠1 ≅ ∠4 𝑔𝑖𝑣𝑒𝑛
𝑚∠1 = 𝑚∠4
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦:
𝐹𝑜𝑟 𝑎𝑛𝑦 𝑁𝑢𝑚𝑏𝑒𝑟 𝑃, 𝑃 = 𝑃
𝑃 𝑖𝑠 𝑟𝑒𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑡𝑜 𝑒𝑞𝑢𝑎𝑙 𝑡ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓
𝑚∠1 𝑎𝑛𝑑 𝑚∠4
𝑚∠2 = 𝑚∠3
𝑆𝑡𝑒𝑝𝑠:
𝑚∠1 + 𝑚∠2 = 𝑚∠3 + 𝑚∠4
𝑚∠1 + 𝑚∠2 − 𝑚∠1 = 𝑚∠3 + 𝑚∠4 − 𝑚∠4
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦:
𝑁 = 𝑀
𝑁 − 𝑃 = 𝑀 − 𝑃
∠2 ≅ ∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
Congruent Complementary Theorem: 𝑃𝑟𝑜𝑣𝑒 ∠2 ≅ ∠3
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠1 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
𝑔𝑖𝑣𝑒𝑛
𝑚∠1 + 𝑚∠2 = 90°
𝑚∠1 + 𝑚∠3 = 90°
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠
90° = 𝑚∠1 + 𝑚∠3 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝑚∠1 + 𝑚∠2 = 𝑚∠1 + 𝑚∠3 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦:
𝐼𝑓 𝑁 = 𝑃 𝑎𝑛𝑑 𝑃 = 𝑀 𝑡ℎ𝑒𝑛, 𝑁 = 𝑀
𝑚∠1 + 𝑚∠2 = 90° = 𝑚∠1 + 𝑚∠3
𝑁 = 𝑃 = 𝑀
𝑚∠2 = 𝑚∠3
𝑆𝑡𝑒𝑝𝑠:
𝑚∠1 + 𝑚∠2 = 𝑚∠1 + 𝑚∠3
𝑚∠1 + 𝑚∠2 − 𝑚∠1 = 𝑚∠1 + 𝑚∠3 − 𝑚∠1
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦:
𝑃 = 𝑚∠1 (𝑃 𝑖𝑠 𝑟𝑒𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑡𝑜 𝑒𝑞𝑢𝑎𝑙 𝑚∠1)
𝑁 = 𝑀
𝑁 − 𝑃 = 𝑀 − 𝑃
∠2 ≅ ∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
Vertical Angles Theorem: 𝑃𝑟𝑜𝑣𝑒 ∠1 ≅ ∠3
∠1 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑛𝑔𝑙𝑒𝑠
∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑛𝑔𝑙𝑒𝑠
𝑔𝑖𝑣𝑒𝑛
𝑁𝑜𝑡𝑒: 𝐻𝑜𝑤 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐴𝑛𝑔𝑙𝑒𝑠?
𝑃𝑎𝑖𝑟 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒𝑠 𝑡ℎ𝑎𝑡 𝑆ℎ𝑎𝑟𝑒 𝑎 𝑉𝑒𝑟𝑡𝑒𝑥,
𝑎𝑟𝑒 𝑁𝑜𝑡 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠, 𝑎𝑛𝑑 𝑎𝑟𝑒 𝑓𝑜𝑟𝑚𝑒𝑑
𝑓𝑟𝑜𝑚 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑛𝑔 𝐿𝑖𝑛𝑒𝑠
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟
𝑚∠1 + 𝑚∠2 = 180°
𝑚∠2 + 𝑚∠3 = 180°
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠
180° = 𝑚∠2 + ∠3 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝑚∠1 + 𝑚∠2 = 𝑚∠2 + 𝑚∠3 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑦
𝐼𝑓 𝑁 = 𝑃 𝑎𝑛𝑑 𝑃 = 𝑀 𝑡ℎ𝑒𝑛, 𝑁 = 𝑀
𝑚∠1 + 𝑚∠2 = 180° = 𝑚∠2 + 𝑚∠3
𝑁 = 𝑃 = 𝑀
𝑚∠1 = 𝑚∠3
𝑆𝑡𝑒𝑝𝑠:
𝑚∠1 + 𝑚∠2 = 𝑚∠2 + 𝑚∠3
𝑚∠1 + 𝑚∠2 − 𝑚∠2 = 𝑚∠2 + 𝑚∠3 − 𝑚∠2
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦:
𝑃 = 𝑚∠2 (𝑃 𝑖𝑠 𝑟𝑒𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑡𝑜 𝑒𝑞𝑢𝑎𝑙 𝑚∠2)
𝑁 = 𝑀
𝑁 − 𝑃 = 𝑀 − 𝑃
∠1 ≅ ∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
Perpendicular Lines Theorems:
Perpendicular Lines Intersect to Form Four
Right Angles Theorem
𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1, ∠2, ∠3, 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒
𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
𝐿𝑖𝑛𝑒𝑠 𝑡 𝑎𝑛𝑑 𝑠 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 (𝑡 ⊥ 𝑠) 𝑔𝑖𝑣𝑒𝑛
∠1 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐿𝑖𝑛𝑒𝑠
𝑚∠1 = 90° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒
𝑚∠1 + 𝑚∠2 = 180° 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
𝑁𝑜𝑡𝑒: 𝑁𝑒𝑒𝑑 𝑡𝑜 𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑦 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑠
𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟
90° + 𝑚∠2 = 180° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 (𝑠𝑢𝑏 90° 𝑓𝑜𝑟 𝑚∠1)
𝑚∠2 = 90° 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
(𝑆𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 90° 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛)
∠2 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒
∠1 ≅ ∠3 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐴𝑛𝑔𝑙𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
𝑁𝑜𝑡𝑒: 𝑁𝑒𝑒𝑑 𝑡𝑜 𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑦 ∠1 𝑎𝑛𝑑 ∠3 𝑎𝑠
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐴𝑛𝑔𝑙𝑒𝑠, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡
𝑚∠1 = 𝑚∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
90° = 𝑚∠3 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 (𝑠𝑢𝑏 90° 𝑓𝑜𝑟 𝑚∠1)
∠3 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒
𝑚∠1 + 𝑚∠4 = 180° 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 (𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟)
90° + 𝑚∠4 = 180° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 (𝑠𝑢𝑏 90° 𝑓𝑜𝑟 𝑚∠1)
𝑚∠4 = 90° 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
(𝑆𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 90° 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛)
∠4 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒
All Right Angles are Congruent Theorem 𝑃𝑟𝑜𝑣𝑒 ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4
𝑡 ⊥ 𝑠 𝑔𝑖𝑣𝑒𝑛
∠1 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
∠2 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
∠3 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
∠4 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐿𝑖𝑛𝑒𝑠 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑡𝑜 𝐹𝑜𝑟𝑚
𝐹𝑜𝑢𝑟 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
𝑚∠1 = 90°
𝑚∠2 = 90°
𝑚∠3 = 90°
𝑚∠4 = 90°
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒
𝑚∠1 = 𝑚∠2 = 𝑚∠3 = 𝑚∠4 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛
∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
Perpendicular Lines Will Form Four
Congruent Adjacent Angles Theorem
𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 ≅ ∠2
𝑡 ⊥ 𝑠 𝑔𝑖𝑣𝑒𝑛
∠1 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
∠2 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
∠3 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
∠4 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐿𝑖𝑛𝑒𝑠 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑡𝑜 𝐹𝑜𝑟𝑚
𝐹𝑜𝑢𝑟 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4 𝐴𝑙𝑙 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
∠3 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
∠4 𝑎𝑛𝑑 ∠1 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
𝑁𝑜𝑡𝑒: 𝐻𝑜𝑤 𝑡𝑜 𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑦 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠?
𝑆ℎ𝑎𝑟𝑒 𝑎 𝑉𝑒𝑟𝑡𝑒𝑥, 𝑆ℎ𝑎𝑟𝑒 𝑎 𝑆𝑖𝑑𝑒,
𝑎𝑛𝑑 𝑎𝑟𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑆𝑎𝑚𝑒 𝑃𝑙𝑎𝑛𝑒
If Two Angles are Both Congruent and
Supplementary, Then Each Angle Will be a
Right Angle Theorem
𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
∠1 ≅ ∠2
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
𝑔𝑖𝑣𝑒𝑛
𝑚∠1 = 𝑚∠2 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
𝑚∠1 + 𝑚∠2 = 180° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠
𝑚∠1 + 𝑚∠1 = 180° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
(𝑠𝑢𝑏 𝑚∠1 𝑓𝑜𝑟 𝑚∠2
2 ∗ 𝑚∠1 = 180° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
(𝑠𝑢𝑏 2 ∗ 𝑚∠1 𝑓𝑜𝑟 𝑚∠1 + 𝑚∠1)
𝑚∠1
2
=
180°
2
𝑚∠1 = 90°
𝐷𝑖𝑣𝑖𝑠𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
∠1 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒
𝑚∠2 = 90° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦
(𝑠𝑢𝑏 𝑚∠2 𝑓𝑜𝑟 𝑚∠1)
∠2 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒
If Two Angles of a Linear Pair are
Congruent, Then They Are Right Angles
Theorem
𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
∠1 ≅ ∠2
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑝𝑎𝑖𝑟
𝑔𝑖𝑣𝑒𝑛
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝐼𝑓 𝑇𝑤𝑜 𝐴𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝐵𝑜𝑡ℎ 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑛𝑑
𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦, 𝑇ℎ𝑒𝑛 𝐸𝑎𝑐ℎ 𝐴𝑛𝑔𝑙𝑒 𝑊𝑖𝑙𝑙
𝑏𝑒 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚

More Related Content

What's hot

engineering statics: distributed forces-1
engineering statics: distributed forces-1engineering statics: distributed forces-1
engineering statics: distributed forces-1
musadoto
 
Wk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasound
Wk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasoundWk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasound
Wk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasound
chris lembalemba
 
engineering statics :force systems
 engineering statics :force systems engineering statics :force systems
engineering statics :force systems
musadoto
 

What's hot (19)

Center of mass_&_rocket_propulsion
Center of mass_&_rocket_propulsionCenter of mass_&_rocket_propulsion
Center of mass_&_rocket_propulsion
 
Mechanics Lec 1
Mechanics Lec 1Mechanics Lec 1
Mechanics Lec 1
 
Reflection refraction
Reflection refractionReflection refraction
Reflection refraction
 
Simple harmonic oscillator - Classical Mechanics
Simple harmonic oscillator - Classical MechanicsSimple harmonic oscillator - Classical Mechanics
Simple harmonic oscillator - Classical Mechanics
 
Civil Engineering structure
Civil Engineering structureCivil Engineering structure
Civil Engineering structure
 
Ce 255 handout
Ce 255 handoutCe 255 handout
Ce 255 handout
 
engineering statics: distributed forces-1
engineering statics: distributed forces-1engineering statics: distributed forces-1
engineering statics: distributed forces-1
 
Teaching Slide1
Teaching Slide1Teaching Slide1
Teaching Slide1
 
Review of Seiberg Witten duality.pptx
Review of Seiberg Witten duality.pptxReview of Seiberg Witten duality.pptx
Review of Seiberg Witten duality.pptx
 
Handbook to ssc je mechanical
Handbook to ssc je mechanical Handbook to ssc je mechanical
Handbook to ssc je mechanical
 
HSC Physics - Torque
HSC Physics - TorqueHSC Physics - Torque
HSC Physics - Torque
 
1 equilbrium
1 equilbrium1 equilbrium
1 equilbrium
 
Wk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasound
Wk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasoundWk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasound
Wk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasound
 
Rigid body solutions
Rigid body solutionsRigid body solutions
Rigid body solutions
 
Balancing of rotating masses
Balancing of rotating massesBalancing of rotating masses
Balancing of rotating masses
 
engineering statics :force systems
 engineering statics :force systems engineering statics :force systems
engineering statics :force systems
 
MECHANICS OF SOLIDS(coplanar concurrent forces)
MECHANICS OF SOLIDS(coplanar concurrent forces)MECHANICS OF SOLIDS(coplanar concurrent forces)
MECHANICS OF SOLIDS(coplanar concurrent forces)
 
Solving trigonometric equations 1
Solving trigonometric equations 1Solving trigonometric equations 1
Solving trigonometric equations 1
 
Centroid & Centre of Gravity
Centroid & Centre of GravityCentroid & Centre of Gravity
Centroid & Centre of Gravity
 

Similar to Geometry Theorems 1 REMC Tutoring

7). mechanical waves (finished)
7). mechanical waves (finished)7). mechanical waves (finished)
7). mechanical waves (finished)
PhysicsLover
 

Similar to Geometry Theorems 1 REMC Tutoring (20)

Differential Geometry for Machine Learning
Differential Geometry for Machine LearningDifferential Geometry for Machine Learning
Differential Geometry for Machine Learning
 
SUEC 高中 Adv Maths (Degree-Radian, Arc length)
SUEC 高中 Adv Maths  (Degree-Radian, Arc length)SUEC 高中 Adv Maths  (Degree-Radian, Arc length)
SUEC 高中 Adv Maths (Degree-Radian, Arc length)
 
07.mdsd_modelado_termicos_liquidos
07.mdsd_modelado_termicos_liquidos07.mdsd_modelado_termicos_liquidos
07.mdsd_modelado_termicos_liquidos
 
FUNDAMENTALS OF HEAT TRANSFER .pdf
FUNDAMENTALS OF HEAT TRANSFER .pdfFUNDAMENTALS OF HEAT TRANSFER .pdf
FUNDAMENTALS OF HEAT TRANSFER .pdf
 
HEAT CONDUCTION DEMYSTIFIED.pdf
HEAT CONDUCTION DEMYSTIFIED.pdfHEAT CONDUCTION DEMYSTIFIED.pdf
HEAT CONDUCTION DEMYSTIFIED.pdf
 
RADIAL HEAT CONDUCTION SOLVED USING THE INTEGRAL EQUATION .pdf
RADIAL HEAT CONDUCTION SOLVED USING THE INTEGRAL EQUATION .pdfRADIAL HEAT CONDUCTION SOLVED USING THE INTEGRAL EQUATION .pdf
RADIAL HEAT CONDUCTION SOLVED USING THE INTEGRAL EQUATION .pdf
 
K to 12 math
K to 12 mathK to 12 math
K to 12 math
 
SUEC 高中 Adv Maths (Trigo Equation) (Part 3)
SUEC 高中 Adv Maths (Trigo Equation) (Part 3)SUEC 高中 Adv Maths (Trigo Equation) (Part 3)
SUEC 高中 Adv Maths (Trigo Equation) (Part 3)
 
Integral method of the Analytic solutions to the heat equation With Experimen...
Integral method of the Analytic solutions to the heat equation With Experimen...Integral method of the Analytic solutions to the heat equation With Experimen...
Integral method of the Analytic solutions to the heat equation With Experimen...
 
Mathematical analysis of non-uniform polyhedra having 2 congruent regular n-g...
Mathematical analysis of non-uniform polyhedra having 2 congruent regular n-g...Mathematical analysis of non-uniform polyhedra having 2 congruent regular n-g...
Mathematical analysis of non-uniform polyhedra having 2 congruent regular n-g...
 
Parallelogram Law Force | Civil Engineering
Parallelogram Law Force | Civil EngineeringParallelogram Law Force | Civil Engineering
Parallelogram Law Force | Civil Engineering
 
7). mechanical waves (finished)
7). mechanical waves (finished)7). mechanical waves (finished)
7). mechanical waves (finished)
 
SUEC 高中 Adv Maths (Trigo Function Part 3)
SUEC 高中 Adv Maths (Trigo Function Part 3)SUEC 高中 Adv Maths (Trigo Function Part 3)
SUEC 高中 Adv Maths (Trigo Function Part 3)
 
TRANSIENT AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION SOLVED ...
TRANSIENT AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION SOLVED ...TRANSIENT AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION SOLVED ...
TRANSIENT AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION SOLVED ...
 
09.sdcd_lugar_geometrico_raices
09.sdcd_lugar_geometrico_raices09.sdcd_lugar_geometrico_raices
09.sdcd_lugar_geometrico_raices
 
SEMI-INFINITE ROD SOLUTION FOR TRANSIENT AND STEADY STATE.pdf
SEMI-INFINITE ROD SOLUTION FOR TRANSIENT AND STEADY STATE.pdfSEMI-INFINITE ROD SOLUTION FOR TRANSIENT AND STEADY STATE.pdf
SEMI-INFINITE ROD SOLUTION FOR TRANSIENT AND STEADY STATE.pdf
 
Equations_3_Industrial Instrumentation - Temperature & Level Measurement Impo...
Equations_3_Industrial Instrumentation - Temperature & Level Measurement Impo...Equations_3_Industrial Instrumentation - Temperature & Level Measurement Impo...
Equations_3_Industrial Instrumentation - Temperature & Level Measurement Impo...
 
Teoria Numérica (Palestra 01)
Teoria Numérica (Palestra 01)Teoria Numérica (Palestra 01)
Teoria Numérica (Palestra 01)
 
Jordan Higher (𝜎, 𝜏)-Centralizer on Prime Ring
Jordan Higher (𝜎, 𝜏)-Centralizer on Prime RingJordan Higher (𝜎, 𝜏)-Centralizer on Prime Ring
Jordan Higher (𝜎, 𝜏)-Centralizer on Prime Ring
 
ANALTICAL SOLUTIONS TO THE HEAT EQUATION USING THE INTEGRAL METHODS.pdf
ANALTICAL SOLUTIONS TO THE HEAT EQUATION USING THE INTEGRAL METHODS.pdfANALTICAL SOLUTIONS TO THE HEAT EQUATION USING THE INTEGRAL METHODS.pdf
ANALTICAL SOLUTIONS TO THE HEAT EQUATION USING THE INTEGRAL METHODS.pdf
 

Recently uploaded

Salient Features of India constitution especially power and functions
Salient Features of India constitution especially power and functionsSalient Features of India constitution especially power and functions
Salient Features of India constitution especially power and functions
KarakKing
 

Recently uploaded (20)

How to Create and Manage Wizard in Odoo 17
How to Create and Manage Wizard in Odoo 17How to Create and Manage Wizard in Odoo 17
How to Create and Manage Wizard in Odoo 17
 
SOC 101 Demonstration of Learning Presentation
SOC 101 Demonstration of Learning PresentationSOC 101 Demonstration of Learning Presentation
SOC 101 Demonstration of Learning Presentation
 
HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx
HMCS Max Bernays Pre-Deployment Brief (May 2024).pptxHMCS Max Bernays Pre-Deployment Brief (May 2024).pptx
HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx
 
Plant propagation: Sexual and Asexual propapagation.pptx
Plant propagation: Sexual and Asexual propapagation.pptxPlant propagation: Sexual and Asexual propapagation.pptx
Plant propagation: Sexual and Asexual propapagation.pptx
 
Accessible Digital Futures project (20/03/2024)
Accessible Digital Futures project (20/03/2024)Accessible Digital Futures project (20/03/2024)
Accessible Digital Futures project (20/03/2024)
 
80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...
80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...
80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...
 
Exploring_the_Narrative_Style_of_Amitav_Ghoshs_Gun_Island.pptx
Exploring_the_Narrative_Style_of_Amitav_Ghoshs_Gun_Island.pptxExploring_the_Narrative_Style_of_Amitav_Ghoshs_Gun_Island.pptx
Exploring_the_Narrative_Style_of_Amitav_Ghoshs_Gun_Island.pptx
 
Jamworks pilot and AI at Jisc (20/03/2024)
Jamworks pilot and AI at Jisc (20/03/2024)Jamworks pilot and AI at Jisc (20/03/2024)
Jamworks pilot and AI at Jisc (20/03/2024)
 
COMMUNICATING NEGATIVE NEWS - APPROACHES .pptx
COMMUNICATING NEGATIVE NEWS - APPROACHES .pptxCOMMUNICATING NEGATIVE NEWS - APPROACHES .pptx
COMMUNICATING NEGATIVE NEWS - APPROACHES .pptx
 
TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...
TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...
TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...
 
Tatlong Kwento ni Lola basyang-1.pdf arts
Tatlong Kwento ni Lola basyang-1.pdf artsTatlong Kwento ni Lola basyang-1.pdf arts
Tatlong Kwento ni Lola basyang-1.pdf arts
 
HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx
HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptxHMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx
HMCS Vancouver Pre-Deployment Brief - May 2024 (Web Version).pptx
 
Graduate Outcomes Presentation Slides - English
Graduate Outcomes Presentation Slides - EnglishGraduate Outcomes Presentation Slides - English
Graduate Outcomes Presentation Slides - English
 
FSB Advising Checklist - Orientation 2024
FSB Advising Checklist - Orientation 2024FSB Advising Checklist - Orientation 2024
FSB Advising Checklist - Orientation 2024
 
NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...
NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...
NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...
 
Fostering Friendships - Enhancing Social Bonds in the Classroom
Fostering Friendships - Enhancing Social Bonds  in the ClassroomFostering Friendships - Enhancing Social Bonds  in the Classroom
Fostering Friendships - Enhancing Social Bonds in the Classroom
 
Basic Intentional Injuries Health Education
Basic Intentional Injuries Health EducationBasic Intentional Injuries Health Education
Basic Intentional Injuries Health Education
 
Sociology 101 Demonstration of Learning Exhibit
Sociology 101 Demonstration of Learning ExhibitSociology 101 Demonstration of Learning Exhibit
Sociology 101 Demonstration of Learning Exhibit
 
Salient Features of India constitution especially power and functions
Salient Features of India constitution especially power and functionsSalient Features of India constitution especially power and functions
Salient Features of India constitution especially power and functions
 
Mehran University Newsletter Vol-X, Issue-I, 2024
Mehran University Newsletter Vol-X, Issue-I, 2024Mehran University Newsletter Vol-X, Issue-I, 2024
Mehran University Newsletter Vol-X, Issue-I, 2024
 

Geometry Theorems 1 REMC Tutoring

  • 1. Theorems Midpoint Theorem 𝐼𝑓 𝐵 𝑖𝑠 𝑡ℎ𝑒 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐶 ̅̅̅̅, 𝑡ℎ𝑒𝑛 𝐴𝐵 ̅̅̅̅ ≅ 𝐵𝐶 ̅̅̅̅ Segment Congruence Theorems Reflexive Symmetric Transitive 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑒 𝐴𝑙𝑔𝑒𝑏𝑟𝑎𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠 𝑜𝑓 𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒, 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐, 𝑎𝑛𝑑 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝐴𝐵 ̅̅̅̅̅ ≅ 𝐴𝐵 ̅̅̅̅ 𝐼𝑓 𝐴𝐵 ̅̅̅̅ ≅ 𝐶𝐷 ̅̅̅̅, 𝑡ℎ𝑒𝑛 𝐶𝐷 ̅̅̅̅ ≅ 𝐴𝐵 ̅̅̅̅ 𝐼𝑓 𝐴𝐵 ̅̅̅̅ ≅ 𝐶𝐷 ̅̅̅̅ 𝑎𝑛𝑑 𝐶𝐷 ̅̅̅̅ ≅ 𝐸𝐹 ̅̅̅̅, 𝑡ℎ𝑒𝑛 𝐴𝐵 ̅̅̅̅ ≅ 𝐸𝐹 ̅̅̅̅ 𝐴𝐵 ̅̅̅̅ ≅ 𝐶𝐷 ̅̅̅̅ 𝑎𝑛𝑑 𝐶𝐷 ̅̅̅̅ ≅ 𝐸𝐹 ̅̅̅̅, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐴𝐵 ̅̅̅̅ ≅ 𝐸𝐹 ̅̅̅̅ (𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑖𝑠 𝑢𝑠𝑒𝑑 𝑤ℎ𝑒𝑛 𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑜𝑡ℎ𝑒𝑟) Supplement Theorem 𝐼𝑓 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑝𝑎𝑖𝑟 𝑖𝑠 𝑓𝑜𝑟𝑚𝑒𝑑 𝑏𝑦 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦. ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 (𝑠𝑢𝑚 𝑜𝑓 𝑚∠1 𝑎𝑛𝑑 𝑚∠2 𝑒𝑞𝑢𝑎𝑙𝑠 180°) Complement Theorem 𝐼𝑓 𝑡ℎ𝑒 𝑢𝑛𝑐𝑜𝑚𝑚𝑜𝑛 𝑠𝑖𝑑𝑒𝑠 𝑓𝑟𝑜𝑚 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝑓𝑜𝑟𝑚 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦. ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 (𝑠𝑢𝑚 𝑜𝑓 𝑚∠1 𝑎𝑛𝑑 𝑚∠2 𝑒𝑞𝑢𝑎𝑙𝑠 90°) (𝑝𝑢𝑟𝑝𝑙𝑒 𝑏𝑜𝑥 𝑠𝑦𝑚𝑏𝑜𝑙 𝑚𝑒𝑎𝑛𝑠 𝑏𝑙𝑎𝑐𝑘 𝑙𝑖𝑛𝑒𝑠 𝑎𝑟𝑒 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 𝑜𝑟 90°)
  • 2. Congruent Supplement Theorem (3 angles) Congruent Supplement Theorem (4 angles) 𝐼𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑑 𝑎𝑛𝑔𝑙𝑒 𝑜𝑛𝑒 𝑖𝑠 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑤𝑖𝑡ℎ 𝑎 𝑡ℎ𝑖𝑟𝑑 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑛𝑔𝑙𝑒. 𝐼𝑓 𝑡𝑤𝑜 𝑠𝑒𝑡𝑠 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒𝑠 𝑐𝑟𝑒𝑎𝑡𝑒 𝑡𝑤𝑜 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑝𝑎𝑖𝑟𝑠, 𝑎𝑛𝑑 𝑜𝑛𝑒 𝑎𝑛𝑔𝑙𝑒 𝑓𝑟𝑜𝑚 𝑒𝑎𝑐ℎ 𝑠𝑒𝑡 𝑖𝑠 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑒𝑡, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑛𝑜𝑡 𝑙𝑎𝑏𝑒𝑙𝑒𝑑 𝑎𝑠 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑙𝑠𝑜. ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠1 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 ∠2 ≅ ∠3 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠3 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠2 ≅ ∠3 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, ∠1 ≅ ∠4 Congruent Complement Theorem 𝐼𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑑 𝑎𝑛𝑔𝑙𝑒 𝑜𝑛𝑒 𝑖𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑤𝑖𝑡ℎ 𝑎 𝑡ℎ𝑖𝑟𝑑 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑛𝑔𝑙𝑒. ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, ∠1 ≅ ∠3 Vertical Angles Theorem 𝐼𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑛𝑔𝑙𝑒𝑠, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡. ∠1 ≅ ∠3 ∠2 ≅ ∠4
  • 3. Right Angle Theorems Perpendicular Lines Intersect to Form Four Right Angles All Right Angles are Congruent Perpendicular Lines Will Form Four Congruent Adjacent Angles If Two Angles are Both Congruent and Supplementary, Then Each Angle Will be a Right Angle If Two Angles of a Linear Pair are Congruent, Then They Are Right Angles 𝐼𝑓 𝑡𝑤𝑜 𝑙𝑖𝑛𝑒𝑠 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟, 𝑡ℎ𝑒𝑛 𝑓𝑜𝑢𝑟 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡. 𝐼𝑓 𝑡𝑤𝑜 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠, 𝑡ℎ𝑒𝑛 𝑎𝑙𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡. 𝐼𝑓 𝑡𝑤𝑜 𝑙𝑖𝑛𝑒𝑠 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑡𝑜 𝑎𝑡 𝑎 90° 𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒𝑛 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑓𝑜𝑟𝑚𝑒𝑑. 𝐼𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑖𝑛 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑠 180°, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑒𝑎𝑐ℎ 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 90° 𝐼𝑓 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑝𝑎𝑖𝑟 𝑖𝑠 𝑓𝑜𝑟𝑚𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡𝑤𝑜 𝑐𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑏𝑜𝑡ℎ 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 90° 𝐹𝑜𝑢𝑟 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑟𝑒𝑎𝑡𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐿𝑖𝑛𝑒𝑠 𝐴𝑙𝑙 𝑡ℎ𝑒 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑒𝑞𝑢𝑎𝑙 90° ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ∠3 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ∠4 𝑎𝑛𝑑 ∠1 𝑎𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑙𝑙 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒 𝑃𝑎𝑖𝑟𝑠 𝑎𝑟𝑒 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝑡ℎ𝑒 𝑝𝑎𝑖𝑟𝑠 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦, 𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 𝑒𝑎𝑐ℎ 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒. 𝐴𝑙𝑙 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒 𝑃𝑎𝑖𝑟𝑠 𝑎𝑟𝑒 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝑐𝑟𝑒𝑎𝑡𝑒 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟𝑠 𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 𝑒𝑎𝑐ℎ 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒. Theorem Proofs Midpoint Theorem: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: 𝐴𝐵 ̅̅̅̅ ≅ 𝐵𝐶 ̅̅̅̅ 𝐵 𝑖𝑠 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐶 ̅̅̅̅ 𝑔𝑖𝑣𝑒𝑛 𝐴𝐵 = 𝐵𝐶 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑎 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐴𝐵 ̅̅̅̅ ≅ 𝐵𝐶 ̅̅̅̅ 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒
  • 4. Segment Congruence Theorems: Reflexive Property: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: 𝐴𝐵 ̅̅̅̅̅ ≅ 𝐴𝐵 ̅̅̅̅ 𝐴𝐵 ̅̅̅̅ ≅ 𝐴𝐵 ̅̅̅̅ 𝑔𝑖𝑣𝑒𝑛 𝐴𝐵 = 𝐴𝐵 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 𝐴𝐵 = 𝐴𝐵 𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒 𝐴𝐵 ̅̅̅̅ ≅ 𝐴𝐵 ̅̅̅̅ 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 Symmetric Property: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: 𝐶𝐷 ̅̅̅̅ ≅ 𝐴𝐵 ̅̅̅̅ 𝐴𝐵 ̅̅̅̅ ≅ 𝐶𝐷 ̅̅̅̅ 𝑔𝑖𝑣𝑒𝑛 𝐴𝐵 = 𝐶𝐷 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 𝐶𝐷 = 𝐴𝐵 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝐶𝐷 ̅̅̅̅ ≅ 𝐴𝐵 ̅̅̅̅ 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 Transitive Property: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: 𝐴𝐵 ̅̅̅̅ ≅ 𝐸𝐹 ̅̅̅̅ 𝐴𝐵 ̅̅̅̅ ≅ 𝐶𝐷 ̅̅̅̅ 𝑎𝑛𝑑 𝐶𝐷 ̅̅̅̅ ≅ 𝐸𝐹 ̅̅̅̅ 𝑔𝑖𝑣𝑒𝑛 𝐴𝐵 = 𝐶𝐷 𝑎𝑛𝑑 𝐶𝐷 = 𝐸𝐹 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 𝐴𝐵 = 𝐸𝐹 𝐴𝐵 = 𝐸𝐹 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑜𝑟, 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 (𝑠𝑢𝑏 𝐴𝐵 𝑓𝑜𝑟 𝐶𝐷) 𝐴𝐵 ̅̅̅̅ ≅ 𝐸𝐹 ̅̅̅̅ 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 Supplement Theorem: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠1 𝑎𝑛𝑑 ∠2 𝑓𝑜𝑟𝑚 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑝𝑎𝑖𝑟 𝑔𝑖𝑣𝑒𝑛 𝑁𝑜𝑡𝑒: 𝐻𝑜𝑤 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑎 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟? 𝑇𝑤𝑜 𝐴𝑛𝑔𝑙𝑒𝑠 𝑡ℎ𝑎𝑡: 𝐴𝑟𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑈𝑛𝑠ℎ𝑎𝑟𝑒𝑑 𝑠𝑖𝑑𝑒𝑠 𝑓𝑜𝑟𝑚 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑟𝑎𝑦𝑠 (180°) 𝑚∠1 + 𝑚∠2 = 180° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠 Complement Theorem: 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 ∠1 𝑎𝑛𝑑 ∠2 𝑓𝑜𝑟𝑚 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝑔𝑖𝑣𝑒𝑛 𝑁𝑜𝑡𝑒: 𝐻𝑜𝑤 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠? 𝑃𝑎𝑖𝑟 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒𝑠 𝑡ℎ𝑎𝑡 𝑆ℎ𝑎𝑟𝑒 𝑎 𝑉𝑒𝑟𝑡𝑒𝑥, 𝑆ℎ𝑎𝑟𝑒 𝑎 𝑆𝑖𝑑𝑒, 𝑎𝑛𝑑 𝑎𝑟𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑆𝑎𝑚𝑒 𝑃𝑙𝑎𝑛𝑒 𝐻𝑜𝑤 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒? 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 = 90° 𝑎𝑛𝑔𝑙𝑒 𝑚∠1 + 𝑚∠2 = 90° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠 Alternative Complement Theorem: ∠1 𝑎𝑛𝑑 ∠2 𝑐𝑟𝑒𝑎𝑡𝑒 ∠𝐴𝐵𝐶, 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦
  • 5. ∠𝐴𝐵𝐶 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 ∠1 𝑎𝑛𝑑 ∠2 𝑓𝑜𝑟𝑚 ∠𝐴𝐵𝐶 𝑔𝑖𝑣𝑒𝑛 𝑚∠𝐴𝐵𝐶 = 90° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝑚∠1 + 𝑚∠2 = 𝑚∠𝐴𝐵𝐶 𝐴𝑛𝑔𝑙𝑒 𝐴𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝑃𝑜𝑠𝑡𝑢𝑙𝑎𝑡𝑒 𝑚∠1 + 𝑚∠2 = 90° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑠𝑢𝑏 90° 𝑖𝑛 𝑓𝑜𝑟 ∠𝐴𝐵𝐶 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠 Congruent Supplement Theorems: Congruent Supplement Theorem (3 angles): 𝑃𝑟𝑜𝑣𝑒 ∠2 ≅ ∠3 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠1 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑔𝑖𝑣𝑒𝑛 𝑚∠1 + 𝑚∠2 = 180° 𝑚∠1 + 𝑚∠3 = 180° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠 180° = 𝑚∠1 + 𝑚∠3 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑚∠1 + 𝑚∠2 = 𝑚∠1 + 𝑚∠3 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦: 𝐼𝑓 𝑁 = 𝑃 𝑎𝑛𝑑 𝑃 = 𝑀 𝑡ℎ𝑒𝑛, 𝑁 = 𝑀 𝑚∠1 + 𝑚∠2 = 180° = 𝑚∠1 + 𝑚∠3 𝑁 = 𝑃 = 𝑀 𝑚∠2 = 𝑚∠3 𝑆𝑡𝑒𝑝𝑠: 𝑚∠1 + 𝑚∠2 = 𝑚∠1 + 𝑚∠3 𝑚∠1 + 𝑚∠2 − 𝑚∠1 = 𝑚∠1 + 𝑚∠3 − 𝑚∠1 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦: 𝑃 = 𝑚∠1 (𝑃 𝑖𝑠 𝑟𝑒𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑡𝑜 𝑒𝑞𝑢𝑎𝑙 𝑚∠1) 𝑁 = 𝑀 𝑁 − 𝑃 = 𝑀 − 𝑃 ∠2 ≅ ∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 Congruent Supplement Theorem (4 angles): 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠2 ≅ ∠3 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑔𝑙𝑒𝑠 ∠3 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑔𝑙𝑒𝑠 ∠1 ≅ ∠4 𝑔𝑖𝑣𝑒𝑛 𝑚∠1 + 𝑚∠2 = 180° 𝑚∠3 + 𝑚∠4 = 180° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠 180° = 𝑚∠3 + 𝑚∠4 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑚∠1 + 𝑚∠2 = 𝑚∠3 + 𝑚∠4 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦: 𝐼𝑓 𝑁 = 𝑃 𝑎𝑛𝑑 𝑃 = 𝑀, 𝑡ℎ𝑒𝑛 𝑁 = 𝑀 𝑚∠1 + 𝑚∠2 = 180° = 𝑚∠3 + 𝑚∠4 𝑁 = 𝑃 = 𝑀 ∠1 ≅ ∠4 𝑔𝑖𝑣𝑒𝑛 𝑚∠1 = 𝑚∠4 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦: 𝐹𝑜𝑟 𝑎𝑛𝑦 𝑁𝑢𝑚𝑏𝑒𝑟 𝑃, 𝑃 = 𝑃 𝑃 𝑖𝑠 𝑟𝑒𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑡𝑜 𝑒𝑞𝑢𝑎𝑙 𝑡ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑚∠1 𝑎𝑛𝑑 𝑚∠4 𝑚∠2 = 𝑚∠3 𝑆𝑡𝑒𝑝𝑠: 𝑚∠1 + 𝑚∠2 = 𝑚∠3 + 𝑚∠4 𝑚∠1 + 𝑚∠2 − 𝑚∠1 = 𝑚∠3 + 𝑚∠4 − 𝑚∠4 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦: 𝑁 = 𝑀 𝑁 − 𝑃 = 𝑀 − 𝑃
  • 6. ∠2 ≅ ∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 Congruent Complementary Theorem: 𝑃𝑟𝑜𝑣𝑒 ∠2 ≅ ∠3 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠1 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑔𝑖𝑣𝑒𝑛 𝑚∠1 + 𝑚∠2 = 90° 𝑚∠1 + 𝑚∠3 = 90° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠 90° = 𝑚∠1 + 𝑚∠3 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑚∠1 + 𝑚∠2 = 𝑚∠1 + 𝑚∠3 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦: 𝐼𝑓 𝑁 = 𝑃 𝑎𝑛𝑑 𝑃 = 𝑀 𝑡ℎ𝑒𝑛, 𝑁 = 𝑀 𝑚∠1 + 𝑚∠2 = 90° = 𝑚∠1 + 𝑚∠3 𝑁 = 𝑃 = 𝑀 𝑚∠2 = 𝑚∠3 𝑆𝑡𝑒𝑝𝑠: 𝑚∠1 + 𝑚∠2 = 𝑚∠1 + 𝑚∠3 𝑚∠1 + 𝑚∠2 − 𝑚∠1 = 𝑚∠1 + 𝑚∠3 − 𝑚∠1 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦: 𝑃 = 𝑚∠1 (𝑃 𝑖𝑠 𝑟𝑒𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑡𝑜 𝑒𝑞𝑢𝑎𝑙 𝑚∠1) 𝑁 = 𝑀 𝑁 − 𝑃 = 𝑀 − 𝑃 ∠2 ≅ ∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 Vertical Angles Theorem: 𝑃𝑟𝑜𝑣𝑒 ∠1 ≅ ∠3 ∠1 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑛𝑔𝑙𝑒𝑠 ∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑛𝑔𝑙𝑒𝑠 𝑔𝑖𝑣𝑒𝑛 𝑁𝑜𝑡𝑒: 𝐻𝑜𝑤 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐴𝑛𝑔𝑙𝑒𝑠? 𝑃𝑎𝑖𝑟 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒𝑠 𝑡ℎ𝑎𝑡 𝑆ℎ𝑎𝑟𝑒 𝑎 𝑉𝑒𝑟𝑡𝑒𝑥, 𝑎𝑟𝑒 𝑁𝑜𝑡 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠, 𝑎𝑛𝑑 𝑎𝑟𝑒 𝑓𝑜𝑟𝑚𝑒𝑑 𝑓𝑟𝑜𝑚 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑛𝑔 𝐿𝑖𝑛𝑒𝑠 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟 𝑚∠1 + 𝑚∠2 = 180° 𝑚∠2 + 𝑚∠3 = 180° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠 180° = 𝑚∠2 + ∠3 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑚∠1 + 𝑚∠2 = 𝑚∠2 + 𝑚∠3 𝑇𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑦 𝐼𝑓 𝑁 = 𝑃 𝑎𝑛𝑑 𝑃 = 𝑀 𝑡ℎ𝑒𝑛, 𝑁 = 𝑀 𝑚∠1 + 𝑚∠2 = 180° = 𝑚∠2 + 𝑚∠3 𝑁 = 𝑃 = 𝑀 𝑚∠1 = 𝑚∠3 𝑆𝑡𝑒𝑝𝑠: 𝑚∠1 + 𝑚∠2 = 𝑚∠2 + 𝑚∠3 𝑚∠1 + 𝑚∠2 − 𝑚∠2 = 𝑚∠2 + 𝑚∠3 − 𝑚∠2 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦: 𝑃 = 𝑚∠2 (𝑃 𝑖𝑠 𝑟𝑒𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑡𝑜 𝑒𝑞𝑢𝑎𝑙 𝑚∠2) 𝑁 = 𝑀 𝑁 − 𝑃 = 𝑀 − 𝑃 ∠1 ≅ ∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
  • 7. Perpendicular Lines Theorems: Perpendicular Lines Intersect to Form Four Right Angles Theorem 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1, ∠2, ∠3, 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝐿𝑖𝑛𝑒𝑠 𝑡 𝑎𝑛𝑑 𝑠 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 (𝑡 ⊥ 𝑠) 𝑔𝑖𝑣𝑒𝑛 ∠1 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐿𝑖𝑛𝑒𝑠 𝑚∠1 = 90° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 𝑚∠1 + 𝑚∠2 = 180° 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑁𝑜𝑡𝑒: 𝑁𝑒𝑒𝑑 𝑡𝑜 𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑦 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑠 𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟 90° + 𝑚∠2 = 180° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 (𝑠𝑢𝑏 90° 𝑓𝑜𝑟 𝑚∠1) 𝑚∠2 = 90° 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 (𝑆𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 90° 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛) ∠2 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 ∠1 ≅ ∠3 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐴𝑛𝑔𝑙𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑁𝑜𝑡𝑒: 𝑁𝑒𝑒𝑑 𝑡𝑜 𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑦 ∠1 𝑎𝑛𝑑 ∠3 𝑎𝑠 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐴𝑛𝑔𝑙𝑒𝑠, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑚∠1 = 𝑚∠3 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 90° = 𝑚∠3 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 (𝑠𝑢𝑏 90° 𝑓𝑜𝑟 𝑚∠1) ∠3 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 𝑚∠1 + 𝑚∠4 = 180° 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 (𝐿𝑖𝑛𝑒𝑎𝑟 𝑃𝑎𝑖𝑟) 90° + 𝑚∠4 = 180° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 (𝑠𝑢𝑏 90° 𝑓𝑜𝑟 𝑚∠1) 𝑚∠4 = 90° 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 (𝑆𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 90° 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛) ∠4 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 All Right Angles are Congruent Theorem 𝑃𝑟𝑜𝑣𝑒 ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4 𝑡 ⊥ 𝑠 𝑔𝑖𝑣𝑒𝑛 ∠1 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 ∠2 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 ∠3 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 ∠4 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐿𝑖𝑛𝑒𝑠 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑡𝑜 𝐹𝑜𝑟𝑚 𝐹𝑜𝑢𝑟 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑚∠1 = 90° 𝑚∠2 = 90° 𝑚∠3 = 90° 𝑚∠4 = 90° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 𝑚∠1 = 𝑚∠2 = 𝑚∠3 = 𝑚∠4 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠
  • 8. Perpendicular Lines Will Form Four Congruent Adjacent Angles Theorem 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 ≅ ∠2 𝑡 ⊥ 𝑠 𝑔𝑖𝑣𝑒𝑛 ∠1 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 ∠2 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 ∠3 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 ∠4 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐿𝑖𝑛𝑒𝑠 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑡𝑜 𝐹𝑜𝑟𝑚 𝐹𝑜𝑢𝑟 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4 𝐴𝑙𝑙 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ∠2 𝑎𝑛𝑑 ∠3 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ∠3 𝑎𝑛𝑑 ∠4 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ∠4 𝑎𝑛𝑑 ∠1 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑁𝑜𝑡𝑒: 𝐻𝑜𝑤 𝑡𝑜 𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑦 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠? 𝑆ℎ𝑎𝑟𝑒 𝑎 𝑉𝑒𝑟𝑡𝑒𝑥, 𝑆ℎ𝑎𝑟𝑒 𝑎 𝑆𝑖𝑑𝑒, 𝑎𝑛𝑑 𝑎𝑟𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑆𝑎𝑚𝑒 𝑃𝑙𝑎𝑛𝑒 If Two Angles are Both Congruent and Supplementary, Then Each Angle Will be a Right Angle Theorem 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 ∠1 ≅ ∠2 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑔𝑖𝑣𝑒𝑛 𝑚∠1 = 𝑚∠2 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 𝑚∠1 + 𝑚∠2 = 180° 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝐴𝑛𝑔𝑙𝑒𝑠 𝑚∠1 + 𝑚∠1 = 180° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 (𝑠𝑢𝑏 𝑚∠1 𝑓𝑜𝑟 𝑚∠2 2 ∗ 𝑚∠1 = 180° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 (𝑠𝑢𝑏 2 ∗ 𝑚∠1 𝑓𝑜𝑟 𝑚∠1 + 𝑚∠1) 𝑚∠1 2 = 180° 2 𝑚∠1 = 90° 𝐷𝑖𝑣𝑖𝑠𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 ∠1 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 𝑚∠2 = 90° 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 (𝑠𝑢𝑏 𝑚∠2 𝑓𝑜𝑟 𝑚∠1) ∠2 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 If Two Angles of a Linear Pair are Congruent, Then They Are Right Angles Theorem 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 ∠1 ≅ ∠2 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑝𝑎𝑖𝑟 𝑔𝑖𝑣𝑒𝑛 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 ∠1 𝑎𝑛𝑑 ∠2 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝐼𝑓 𝑇𝑤𝑜 𝐴𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝐵𝑜𝑡ℎ 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑎𝑛𝑑 𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦, 𝑇ℎ𝑒𝑛 𝐸𝑎𝑐ℎ 𝐴𝑛𝑔𝑙𝑒 𝑊𝑖𝑙𝑙 𝑏𝑒 𝑎 𝑅𝑖𝑔ℎ𝑡 𝐴𝑛𝑔𝑙𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚