Question

In triangle abc angle a is obtuse pb perpendicular pc qc is perpendicular to qb prove that ab*aq=ac*ap

Answer 1

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 BQ ⇒ 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]