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)
BC^2 =(AC x CP + AB x BQ)
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∠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]
∠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.