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- 1. 𝑷𝑹𝑷𝑶𝑹𝑪𝑰𝑶𝑵𝑨𝑳𝑰𝑫𝑨𝑫 𝑫𝑬 𝑺𝑬𝑮𝑴𝑬𝑵𝑻𝑶𝑺 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑫𝑬 𝑻𝑯𝑨𝑳𝑬𝑺 𝑆𝑖: 𝑙1 ∥ 𝑙2 ∥ 𝑙3 𝑙1 𝑙2 𝑙3 𝑎 𝑏 𝑚 𝑛 𝑎 𝑏 = 𝑚 𝑛 𝑂𝑏𝑠: 𝑆𝑖: 𝐴𝐶 ∥ 𝑀𝑁 𝐴 𝐵 𝐶 𝑀 𝑁 𝑎 𝑏 𝑚 𝑛 𝑎 𝑏 = 𝑚 𝑛 𝑆𝑖: 𝐴𝐶 ∥ 𝑀𝑁 𝐴 𝐶 𝑀 𝑁 𝑎 𝑏 𝑚 𝑛 𝑎 𝑏 = 𝑚 𝑛 𝛼 𝛼 𝑎 𝑏 𝑚 𝑛 𝐵𝑖𝑠𝑒𝑐𝑡𝑟𝑖𝑧 𝐸𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝜃 𝜃 𝑏 𝑎 𝑚 𝑛 𝑎 𝑏 = 𝑚 𝑛 𝑎 𝑏 = 𝑚 𝑛 𝜃 𝜃 𝛼𝛼 𝐴 𝐵 𝐶 𝐷 𝑂𝑏𝑠: 𝐴, 𝐵, 𝐶 𝑦 𝐷 𝑓𝑜𝑟𝑚𝑎𝑛 𝑢𝑛𝑎 𝑐𝑢𝑎𝑡𝑒𝑟𝑛𝑎 𝑎𝑟𝑚𝑜𝑛𝑖𝑐𝑎. 𝑎 𝑏 𝑐 𝑑 𝑎 𝑏 = 𝑑 𝑐 𝐵𝑖𝑠𝑒𝑐𝑡𝑟𝑖𝑧 𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑫𝑬 𝑩𝑰𝑺𝑬𝑪𝑻𝑹𝑰𝒁
- 2. 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑫𝑬𝑳 𝑰𝑵𝑪𝑬𝑵𝑻𝑹𝑶 𝑐 𝛼 𝛼 𝐼 𝑥 𝑦 𝑎 𝑏 𝑥 𝑦 = 𝑎 + 𝑏 𝑐 𝐼: 𝐼𝑛𝑐𝑒𝑛𝑡𝑟𝑜 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑫𝑬 𝑪𝑬𝑩𝑨 𝑎 𝑚 𝑏 𝑛 𝑐 𝑙 𝑎𝑏𝑐 = 𝑚𝑛𝑙 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑫𝑬 𝑴𝑬𝑵𝑬𝑳𝑨𝑶 𝑎 𝑏 𝑐 𝑚 𝑛 𝑙 𝑎𝑏𝑐 = 𝑚𝑛𝑙 𝑂𝑏𝑠: 𝐴 𝐵 𝐶 𝐷 𝐴, 𝐵, 𝐶 𝑦 𝐷 𝑓𝑜𝑟𝑚𝑎𝑛 𝑢𝑛𝑎 𝑐𝑢𝑎𝑡𝑒𝑟𝑛𝑎 𝑎𝑟𝑚𝑜𝑛𝑖𝑐𝑎. 𝑎 𝑏 𝑐 𝑑 𝑎 𝑏 = 𝑑 𝑐
- 3. 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 𝐴 𝐵 𝐶 𝑀 𝑄 𝐹 𝑥 12 𝐺 𝐺: 𝐵𝑎𝑟𝑖𝑐𝑒𝑛𝑡𝑟𝑜 𝑑𝑒𝑙 ∆𝐴𝐵𝐶: 𝐺 𝐺: 𝐵𝑎𝑟𝑖𝑐𝑒𝑛𝑡𝑟𝑜 𝑎 2𝑎 𝐺: 𝐵𝑎𝑟𝑖𝑐𝑒𝑛𝑡𝑟𝑜 𝐺 𝑎 2𝑎 𝑃𝑜𝑟 𝑡𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑡ℎ𝑎𝑙𝑒𝑠: 2𝑎 𝑎 = 12 𝑥 𝑥 = 6
- 4. 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 12 4 5 𝑥 4𝑘 5𝑘 𝑃𝑜𝑟 𝑡𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑡ℎ𝑎𝑙𝑒𝑠: 12 𝑥 = 4𝑘 5𝑘 𝑥 = 15 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 54 36 70 𝛼 𝛼 2𝑘 3𝑘 𝐷𝑒𝑙 𝑔𝑟𝑎𝑓𝑖𝑐𝑜: 5𝑘 = 70 𝑘 = 14 28 𝑦 42
- 5. 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 𝐴 𝐵 𝐶 6 8 10 𝐻 𝑃 𝑀 𝑥 𝛼 𝛼 𝜃 𝜃 𝐸𝑙 ∆𝑀𝐵𝑃 𝑒𝑠 𝑖𝑠𝑜𝑠𝑐𝑒𝑙𝑒𝑠: 𝑥 (8 − 𝑥) 𝑃𝑜𝑟 𝑡𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑙𝑎 𝑏𝑖𝑠𝑒𝑐𝑡𝑟𝑖𝑧 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟: 6 10 = 𝑥 8 − 𝑥 3 5 24 − 3𝑥 = 5𝑥 𝑥 = 3 𝜃
- 6. 𝐴 𝐵 𝐶 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 10 6 𝐷 𝛼 𝛼 𝐸 𝑥 5𝑘 3𝑘 𝑃𝑜𝑟 𝑡𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑚𝑒𝑛𝑒𝑙𝑎𝑜: (6 − 𝑥) 𝑎 𝑎 6 − 𝑥 𝑎 5𝑘 = 𝑥(𝑎)(8𝑘) 30 − 5𝑥 = 8𝑥 𝑥 = 30 13 𝐷 𝐵 𝐶 𝐴 5𝑘 3𝑘 𝑎 𝑎 𝑥 (6 − 𝑥) 𝐸
- 7. 𝑺𝑬𝑴𝑬𝑱𝑨𝑵𝒁𝑨 𝑫𝑬 𝑻𝑹𝑰𝑨𝑵𝑮𝑼𝑳𝑶𝑺 𝐷𝑜𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑜𝑠 𝑠𝑒𝑟𝑎𝑛 𝑠𝑒𝑚𝑒𝑗𝑎𝑛𝑡𝑒𝑠 𝑠𝑖 𝑡𝑖𝑒𝑛𝑒𝑛 𝑙𝑎 𝑚𝑖𝑠𝑚𝑎 𝑓𝑜𝑟𝑚𝑎. 𝑀𝑖𝑠𝑚𝑎 𝐹𝑜𝑟𝑚𝑎 𝐴𝑛𝑔𝑢𝑙𝑜𝑠 𝑑𝑒 𝑖𝑔𝑢𝑎𝑙 𝑚𝑒𝑑𝑖𝑑𝑎. 𝐴 𝐵 𝐶 𝛼 𝛽 𝜃 𝑀 𝑁 𝐿 𝛼 𝛽 𝜃 ~ ∆𝐴𝐵𝐶~∆𝑀𝑁𝐿 𝑎 𝑏 𝑐 𝑚 𝑛 𝑙 𝐿𝑎𝑑𝑜𝑠 𝐻𝑜𝑚𝑜𝑙𝑜𝑔𝑜𝑠 𝑎 𝑦 𝑚 𝑏 𝑦 𝑛 𝑐 𝑦 𝑙 𝑎 𝑚 = 𝑏 𝑛 = 𝑐 𝑙 𝑪𝑨𝑺𝑶𝑺 𝑫𝑬 𝑺𝑬𝑴𝑬𝑱𝑨𝑵𝒁𝑨 𝛼 𝛼 𝜃 𝜃 ~ 𝑪𝑨𝑺𝑶 I 𝑏 𝑏𝑘 ~ 𝑪𝑨𝑺𝑶 II 𝑎 𝑎𝑘 𝑐 𝑐𝑘 𝑏 𝑏𝑘 ~ 𝑪𝑨𝑺𝑶 III 𝑎 𝑎𝑘 𝛼 𝛼
- 8. 𝑻𝑬𝑶𝑹𝑬𝑴𝑨 𝑫𝑬 𝑳𝑨 𝑨𝑵𝑻𝑰𝑷𝑨𝑹𝑨𝑳𝑬𝑳𝑨 𝛼 𝛼 𝑥 𝑎 𝑏 𝑥2 = a. b 𝐴 𝐵 𝐶 𝐷 𝑀 𝑁 𝑆𝑖: 𝐵𝐶 ∥ 𝐴𝐷 ∥ 𝑀𝑁 𝑥 𝑎 𝑏 𝑚 𝑛 𝑥 = 𝑎. 𝑚 + 𝑏. 𝑛 𝑚 + 𝑛 𝑂𝑏𝑠1: 𝑂𝑏𝑠2: 𝑎 𝑏 𝑥 𝑥 = 𝑎𝑏 𝑎 + 𝑏 𝑂𝑏𝑠3: 𝑥 𝑥 𝑏 ℎ 𝑥 = 𝑏ℎ 𝑏 + ℎ
- 9. 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 𝑎 2𝑎 3𝑎 5 𝑥 2𝑏 𝑏 3𝑏 𝐸𝑙 ∆𝐷𝐸𝐹~∆𝐶𝐹𝐻: 𝛼 𝛼 𝜃 𝜃 𝛽 𝛽 𝑥 5 = 5𝑏 𝑏 𝑥 = 25 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 4 5 𝛼 𝜃 𝛼 = 𝜃 𝛼 𝛼 𝑃𝑜𝑟 𝑡𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑙𝑎 𝑎𝑛𝑡𝑖𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑎: 𝑥 𝑥2 = 9(4) 𝑥 = 6
- 10. 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 3 2 4 12 𝑥 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 4,5 8 𝑥 𝛼 𝛼 9 𝐸𝑙 ∆𝑀𝐵𝐻~∆𝐻𝑄𝐵: 𝑄 sen 𝛼 = 𝑥 9 = 8 𝑥 𝑥 = 6 2 3 ∆𝐴𝑁𝐶~∆𝐹𝐸𝐶 y ∆𝐷𝐵𝐸~∆𝑀𝐵𝑁: 2𝑥 3𝑥 6𝑥 = 12 𝑥 = 2 𝐴 𝑁 𝐶 3 2 𝐹 𝐸 𝑥 2𝑥 3𝑥 𝐷 𝐵 𝐸 𝑀 𝑁 𝑥 3 4
- 11. 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 1 3 𝛼 𝛼 𝐸𝑙 ∆𝐵𝐹𝐷~∆𝐶𝐻𝐷: 𝑅 𝜃 8 𝛼 4 𝑀 sen 𝛼 = 8 𝑅 = 1 4 𝑅 = 32 𝑅 𝐵 𝐶 𝑀 8 8
- 12. 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 6 4 12 𝑥 𝛼 𝛼 𝜃 𝐸𝑙 ∆𝐴𝑄𝐶~∆𝐴𝐵𝑄 𝑎 2𝑎 2𝑎 𝛽 𝛽 𝜔 𝐸𝑙 ∆𝐴𝐶𝑃~∆𝐴𝑃𝐵 𝑥 4 = 2𝑎 𝑎 𝑥 = 8 𝐴 𝑄 𝐶 𝜃 𝛼 12 𝐵 𝑄 𝐴 𝜃 𝛼 6 ~ 𝑎 2𝑎
- 13. 𝑚 𝑺𝒐𝒍𝒖𝒄𝒊𝒐𝒏: 𝑛 𝑓 𝑥 𝑥 𝑥 𝑎 𝑃𝑜𝑟 𝑡𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑡ℎ𝑎𝑙𝑒𝑠: 𝑎 𝑥 = 𝑥 𝑓 𝑎 = 𝑥2 𝑓 𝑃𝑜𝑟 𝑠𝑒𝑚𝑒𝑗𝑎𝑛𝑧𝑎: 𝑎 𝑚 = 𝑛 𝑥 𝑥2 𝑓𝑚 = 𝑛 𝑥 𝑥3 = 𝑚𝑛𝑓 𝑥3 = 27 𝑥 = 3