Question

In triangle ABC , angle A is obtuse , PB perpendicular AC and QC perpendicular AB . Prove that. BC^2 (square) = (AC×CP+AB×BQ)

Answer 1

Here is the answer to your question.


Given: ΔABC,∠BAC is obtuse. PB⊥CA and CQ⊥AB
To prove: (1) AB �AQ = AC �AP
(2) BC2 = AC �CP + AB �BQ
Proof: In ∆ACQ and ∆ABP,
∠CAQ = ∠BAP (Vertically opposite angles)
∠Q = ∠P (90�)





In right ∆BCQ,
⇒ BC2 = CQ2 + QB2
⇒ BC2 = CQ2 + (QA + AB)2
⇒ BC2 = CQ2 + QA2 + AB2 + 2QA �AB
⇒ BC2 = AC2 + AB2 + QA �AB + QA �AB ( In right ∆ ACQ, CQ2 + QA2 = AC2)
⇒ BC2 = AC2 + AB2 + QA �AB + AC �AP (Using (1))
⇒ BC2 = AC (AC + AP) + AB (AB + QA)
⇒ BC2 = AC *CP + AB *BQ