Question

In triangle ABC angle A =60degree prove that BC*BC=AB*AB AC*AC-AB*AC

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 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]