This document discusses the image of triangle ABC under several transformations. It is given that triangle ABC is subjected to transformations including a reflection over line PQ, followed by translations along lines t and s, and a rotation of angle φ1 about point A. The document proves directly and indirectly that the final image of triangle ABC is triangle A6B6C6, which is a rotation of triangle ABC by an angle of φ7 about point I. It provides the conditions used at each step of the indirect proof.

2. Tugas Akhir Geometri Transformasi
Dosen Pengampu: Drs. Suhito, M.Pd.
Oleh:
1. Shinta Puja Chaterine (4101415002)
2. Rosa Ary Ardhini (4101415019)
3. Hana Masfufah (4101415034)
4. Indah Wulansari (4101415056)
5. Puji Lestari (4101415105)

3. BAYANGAN DARI DAERAH (REGION) SEGITIGA ABC OLEH
𝐺 𝑃𝑄, ∘ 𝑀𝑡 ∘ 𝑀𝑠 ∘ 𝐺 𝑘𝑙 ∘ 𝑅 𝐴𝜑1
∘ 𝑆 𝐵
𝑮 𝑷𝑸, ∘ 𝑴 𝒕 ∘ 𝑴 𝒔 ∘ 𝑮 𝒌𝒍
∘ 𝑹 𝑨𝝋 𝟏
∘ 𝑺 𝑩 △ 𝑨𝑩𝑪 = 𝑹 𝑰,𝑸 𝟕
△ 𝑨𝑩𝑪Menunjukkan bahwa:
Diketahui :
1. 𝒂 ⊥ 𝒃, 𝒂 ∩ 𝒃 = {𝑩}
2. 𝒄 ∩ 𝒅 = 𝑨 , 𝒎 ∠𝒄, 𝒅 =
𝟏
𝟐
𝝋 𝟏
3. 𝑲𝑳 sembarang,
4. Garis s dan t sembarang,
5. 𝑷𝑸 ∕∕ 𝒕
Lihat
Kondisi
Awal

4. a
B
d
s
c
𝑔
P
Q
𝑓
A 1
2
𝜑1

5. GAMBAR BUKTI LANGSUNG
NEXT

6. a
B
d
s
c
𝑔
P
Q
𝑓
A 1
2
𝜑1
A
B
C
𝐴1
𝐵1
𝐶1
𝐴2
𝐵2
𝐶2
𝐴3
𝐶3
𝐵3
𝐴4
𝐵4
𝐶4
𝐴5
𝐵5
𝐶5
𝐴6
𝐵6
𝐶6

7. BUKTI LANGSUNG
𝑮 𝑷𝑸, ∘ 𝑴 𝒕 ∘ 𝑴 𝒔 ∘ 𝑮 𝒌𝒍
∘ 𝑹 𝑨𝝋 𝟏
∘ 𝑺 𝑩 △ 𝑨𝑩𝑪
= (𝑮 𝑷𝑸, ∘ 𝑴 𝒕) ∘ 𝑴 𝒔 ∘ 𝑮 𝒌𝒍
∘ 𝑹 𝑨𝝋 𝟏
△ 𝑨 𝟏 𝑩 𝟏 𝑪 𝟏
= (𝑮 𝑷𝑸, ∘ 𝑴 𝒕) ∘ 𝑴 𝒔 ∘ 𝑮 𝒌𝒍
△ 𝑨 𝟐 𝑩 𝟐 𝑪 𝟐
= (𝑮 𝑷𝑸, ∘ 𝑴 𝒕) ∘ 𝑴 𝒔 △ 𝑨 𝟑 𝑩 𝟑 𝑪 𝟑
= (𝑮 𝑷𝑸, ∘ 𝑴 𝒕) △ 𝑨 𝟒 𝑩 𝟒 𝑪 𝟒
= 𝑮 𝑷𝑸, △ 𝑨 𝟓 𝑩 𝟓 𝑪 𝟓
=△ 𝑨 𝟔 𝑩 𝟔 𝑪 𝟔

8. BAYANGAN DARI DAERAH (REGION) SEGITIGA ABC OLEH
𝐺 𝑃𝑄, ∘ 𝑀𝑡 ∘ 𝑀𝑠 ∘ 𝐺 𝑘𝑙 ∘ 𝑅 𝐴𝜑1
∘ 𝑆 𝐵
𝑮 𝑷𝑸, ∘ 𝑴 𝒕 ∘ 𝑴 𝒔 ∘ 𝑮 𝒌𝒍
∘ 𝑹 𝑨𝝋 𝟏
∘ 𝑺 𝑩 △ 𝑨𝑩𝑪 = 𝑹 𝑰,𝑸 𝟕
△ 𝑨𝑩𝑪Menunjukkan bahwa:
Diketahui :
1. 𝒂 ⊥ 𝒃, 𝒂 ∩ 𝒃 = {𝑩}
2. 𝒄 ∩ 𝒅 = 𝑨 , 𝒎 ∠𝒄, 𝒅 =
𝟏
𝟐
𝝋 𝟏
3. 𝑲𝑳 sembarang,
4. Garis s dan t sembarang,
5. 𝑷𝑸 ∕∕ 𝒕
Lihat
Kondisi
Awal

9. BUKTI TAK LANGSUNG
𝑮 𝑷𝑸, ∘ 𝑴 𝒕 ∘ 𝑴 𝒔 ∘ 𝑮 𝒌𝒍
∘ 𝑹 𝑨𝝋 𝟏
∘ 𝑺 𝑩
= 𝑴 𝒉 ∘ 𝑴 𝒈 ∘ 𝑴 𝒕 ∘ 𝑴 𝒔 ∘ 𝑴 𝒇 ∘ 𝑴 𝒆 ∘ 𝑴 𝒅 ∘ 𝑴 𝒄 ∘ 𝑴 𝒃 ∘ 𝑴 𝒂 __(A)
= 𝑴 𝒉 ∘ 𝑴 𝒈 ∘ 𝑴 𝒕 ∘ 𝑴 𝒔 ∘ 𝑴 𝒇 ∘ 𝑴 𝒆 ∘ 𝑴 𝒅 ∘ 𝑴 𝒄 ∘ 𝑴 𝒃 ∘ 𝑴 𝒂 _______(B)
= 𝑴 𝒉 ∘ 𝑺 𝒄 ∘ 𝑹 𝑫,𝝋 𝟐
∘ 𝑴 𝒆 ∘ 𝑹 𝑨,𝝋 𝟏
∘ 𝑺 𝑩_______________(B)
= 𝑴 𝒉 ∘ 𝑺 𝒄 ∘ 𝑹 𝑫,𝝋 𝟐
∘ 𝑴 𝒆 ∘ 𝑴𝒊 ∘ 𝑴𝒋 ∘ 𝑴𝒋 ∘ 𝑴 𝒌 _________(C)
= 𝑴 𝒉 ∘ 𝑺 𝒄 ∘ 𝑹 𝑫,𝝋 𝟐
∘ 𝑴 𝒆 ∘ 𝑴𝒊 ∘ 𝑰 ∘ 𝑴 𝒌________________(D)
= 𝑴 𝒉 ∘ 𝑺 𝒄 ∘ 𝑹 𝑫,𝝋 𝟐
∘ 𝑹 𝑬,𝝋 𝟑
∘ 𝑴 𝒌_________________(D)
= 𝑴 𝒉 ∘ 𝑺 𝒄 ∘ 𝑴𝒍 ∘ 𝑴 𝒎 ∘ 𝑴 𝒎 ∘ 𝑴 𝒏 ∘ 𝑴 𝒌 ___________ (E)
= 𝑴 𝒉 ∘ 𝑺 𝒄 ∘ 𝑴𝒍 ∘ 𝑰 ∘ 𝑴 𝒏 ∘ 𝑴 𝒌 ___________ (F)
= 𝑴 𝒉 ∘ 𝑺 𝒄 ∘ 𝑹 𝑭,𝝋 𝟒
∘ 𝑴 𝒌 ___________ (F)
= 𝑴 𝒉 ∘ 𝑴 𝒐 ∘ 𝑴 𝒑 ∘ 𝑴 𝒑 ∘ 𝑴 𝒒 ∘ 𝑴 𝒌 ___________ (G)

10. = 𝑴 𝒉 ∘ 𝑴 𝒐 ∘ 𝑰 ∘ 𝑴 𝒒 ∘ 𝑴 𝒌 ___________ (H)
= 𝑹 𝑮,𝝋 𝟓
∘ 𝑹 𝑯,𝝋 𝟔
___________ (H)
= 𝑴 𝒖 ∘ 𝑴 𝒓 ∘ 𝑴 𝒓 ∘ 𝑴 𝒗 ___________ (I)
= 𝑴 𝒖 ∘ 𝑰 ∘ 𝑴 𝒗 ___________ (I)
= 𝑴 𝒖 ∘ 𝑴 𝒗 ___________ (J)
= 𝑹 𝑰,𝝋 𝟕
___________ (J)
Jadi ,
𝑮 𝑷𝑸, ∘ 𝑴 𝒕 ∘ 𝑴 𝒔 ∘ 𝑮 𝒌𝒍 ∘ 𝑹 𝑨𝝋 𝟏
∘ 𝑺 𝑩 △ 𝑨𝑩𝑪
= 𝑹 𝑰,𝑸 𝟕
△ 𝑨𝑩𝑪
BUKTI TAK LANGSUNG
Lanjutan...

11. Syarat pada Bukti Tak Langsung
(A) 1) 𝑎 ⊥ 𝑏, 𝑎 ∩ 𝑏 = 𝐵
2) 𝑐 ∩ 𝑑 = 𝐴 , 𝑚 ∠𝑐, 𝑑 =
1
2
𝜑1
3) 𝑒//𝑓, 𝑒 ⊥ 𝐾𝐿, 𝑑 𝑒,𝑓 =
1
2
𝐾𝐿
4) t//𝑃𝑄
5) 𝑔//ℎ, 𝑔 ⊥ 𝑃𝑄, 𝑑 𝑔,ℎ =
1
2
𝑃𝑄
(B) 1) 𝑓 ∩ 𝑠 = 𝐷 , 𝑚 ∠𝑓, 𝑠 =
1
2
𝜑2
2) 𝑡 ∩ 𝑔 = 𝐶 , 𝑡 ⊥ 𝑔
(C) 1) 𝑘 ∩ 𝑗 = 𝐵 , 𝐾 ⊥ 𝐽
2) 𝑗 ∩ 𝑖 = 𝐴 , 𝑚 ∠𝑗, 𝑖 =
1
2
𝜑1
(D) 1) 𝑖 ∩ 𝑒 = 𝐸 , 𝑚 ∠𝑖, 𝑒 =
1
2
𝜑3
(E) 1) 𝑚 ∩ 𝑙 = 𝐷 , 𝑚 ∠𝑚, 𝑙 =
1
2
𝜑2
2) 𝑛 ∩ 𝑚 = 𝐸 , 𝑚 ∠𝑛, 𝑚 =
1
2
𝜑3
(F) 1) 𝑛 ∩ 𝑙 = {𝐹}
2) 𝑚 ∠𝑛, 𝑙 =
1
2
𝜑4
(G) 1) 𝑝 ⊥ 𝑜, 𝑝 ∩ 𝑜 = {𝐶}
2) 𝑞 ∩ 𝑝 = 𝐹 ,
3) 𝑚 ∠𝑞, 𝑝 =
1
2
𝜑4
(H) 1) 𝑘 ∩ 𝑞 = 𝐻 ,
2) 𝑚 ∠𝑘, 𝑞 =
1
2
𝜑6
3) 𝑜 ∩ ℎ = 𝐺 ,
4) 𝑚 ∠𝑜, ℎ =
1
2
𝜑5
(I) 1) 𝑣 ∩ 𝑟 = 𝐻 ,
2) 𝑚 ∠𝑣, 𝑟 =
1
2
𝜑6
3) 𝑟 ∩ 𝑢 = 𝐺 ,
4) 𝑚 ∠𝑟, 𝑢 =
1
2
𝜑5
(J) 1) 𝑣 ∩ 𝑢 = 𝐼 ,
2) 𝑚 ∠𝑣, 𝑢 =
1
2
𝜑7

12. GAMBAR BUKTI TAK LANGSUNG
NEXT

13. a
B
d
s
c
c
𝑔
P
Q
𝑓
A 1
2
𝜑1
A
B
C
𝐴1
𝐵1
𝐶1
𝐴2
𝐵2
𝐶2
𝐴3
𝐶3
𝐵3
𝐴4
𝐵4
𝐶4
𝐴5
𝐵5
𝐶5
𝐴6
𝐵6
𝐶6

14. a
d
s
c
𝑔
P
Q
𝑓
A
A
B
C
𝐴6
𝐵6
𝐶6
C
D
1
2
𝜑1
B
1
2
𝜑1
1
2
𝜑4
1
2
𝜑4
E
F
o
H
G
I
1
2
𝜑6
1
2
𝜑6