Question

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Answer 1

Answer:

Step-by-step explanation:

Here is the answer to your question.

∠P = ∠Q = 90°

∠PAB = ∠QAC [Vertically opposite angles]

⇒ ΔBPA ~ ΔAQC [AAA similarity criterion]

Consider, right angled triangle BCQ

⇒ BC2 = CQ2 + BQ2 [By Pythagoras theorem]

⇒ BC2 = CQ2 + (AB + AQ)2   [Since BQ = AB + AQ]

⇒ BC2 = [CQ2 + AQ2] + AB2 + 2AB × AQ à (2)

In right ∆ACQ, CQ2 + AQ2 = AC2 [By Pythagoras theorem]

Hence equation (2) becomes,

⇒ BC2 = AC2 + AB2 + AB × AQ + AB × AQ

⇒ BC2 = AC2 + AB2 + AB × AQ + AP × AC [From (1)]

⇒ BC2 = AC2 + AP × AC + AB2 + AB × AQ

⇒ BC2 = AC (AC + AP) + AB (AB + AQ)

⇒ BC2 = AC × CP + AB × BQ  [From the figure]

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