Question

In triangle ABC, angle A is obtuse, PB is perpendicular to AC and QC is perpendicular to AB. Prove that

(i) AB X AQ = AC X AP

(ii) BC2 = (AC X CP + AB X BQ)

Answer 1

you can easily google it man and it is even less data consuming than doing this and waiting for the answer mate I hope you understand thank you mate.i would also like to thank you for your question it is kinda easy man.

Answer 2

Answer:

Given in ΔABC, ∠A is obtuse and PB⊥ AC and QC⊥ AB Consider ΔBPA and ΔAQC ∠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]