Question

In triangle ABC, if < A is obtuse,PB perpendicular to AC and QC perpendicular AB .Prove that:
BC^2 =(AC x CP + AB x BQ)

Answer 1

∠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]

Answer 2

∠P=∠Q=90

o

∠PAB=∠QAC [Vertically opposite angles]

∴ △BPA∼△AQC [AAA similarity criterion]

⇒

AC

AB

=

AQ

AP

∴ AB×AQ=AC×AP --- ( 1 ) [ Hence proved ]

Consider, right angled △BCQ

⇒ BC

2

=CQ

2

+BQ

2

[By Pythagoras theorem]

⇒ BC

2

=CQ

2

+(AB+AQ)

2

[Since BQ = AB + AQ]

⇒ BC

2

=[CQ

2

+AQ

2

]+AB

2

+2AB×AQ ----- (2)

In right △ACQ, CQ

2

+AQ

2

=AC

2

[By Pythagoras theorem]

Hence equation (2) becomes,

⇒ BC

2

=AC

2

+AB

2

+AB×AQ+AB×AQ

⇒ BC

2

=AC

2

+AB

2

+AB×AQ+AP×AC [From (1)]

⇒ BC

2

=AC

2

+AP×AC+AB

2

+AB×AQ

⇒ BC

2

=AC(AC+AP)+AB(AB+AQ)

⇒ BC

2

=AC×CP+AB×BQ [From the figure]

Hence Proved.