In a parallelogram ABCD, AP and CQ are drawn perpendiculars from vertices A and C on diagonal BD. Prove that triangle APB = CQD
Answers
Answer:
Answer:Proof : (i) In ∆APB and ∆CQD,
Answer:Proof : (i) In ∆APB and ∆CQD, we have ∠ABP = ∠CDQ [Alternate angles]
Answer:Proof : (i) In ∆APB and ∆CQD, we have ∠ABP = ∠CDQ [Alternate angles] AB = CD [Opposite sides of a parallelogram]
Answer:Proof : (i) In ∆APB and ∆CQD, we have ∠ABP = ∠CDQ [Alternate angles] AB = CD [Opposite sides of a parallelogram] ∠APB = ∠CQD [Each = 90°]
Answer:Proof : (i) In ∆APB and ∆CQD, we have ∠ABP = ∠CDQ [Alternate angles] AB = CD [Opposite sides of a parallelogram] ∠APB = ∠CQD [Each = 90°] ∴ ∆APB ≅ ∆CQD [ASA congruence]
Answer:Proof : (i) In ∆APB and ∆CQD, we have ∠ABP = ∠CDQ [Alternate angles] AB = CD [Opposite sides of a parallelogram] ∠APB = ∠CQD [Each = 90°] ∴ ∆APB ≅ ∆CQD [ASA congruence] (ii) So, AP = CQ [CPCT]
Answer:
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