ABC and DBC are two right triangles Prove that AP×PC
= BP×PD
Answers
Answer:
Given,
Two right triangles ABC & DBC and they are drawn on the same hypotenuse BC & on the same sides BC. Ref. image
AC & BD intersect each other. at 'P'. As per given data
Also given, to prove that
AP×PC=BP×PD
as both are right angle triangles, let us use Pythagorean's theorem for the given triangles.
as per fig (a)
From ΔCDP we get,
CD
2
=CP
2
−DP
2
(∵ from fig (a) we get, CP=CA+AP DP=DB+BP)
=(CA+AP)
2
−(DB+BP)
2
[∵(a+b)
2
=a
2
+2ab+b
2
]
CD
2
=CA
2
+AP
2
+2CA.AP−[DB
2
+BP
2
+2DB.BP]
CD
2
=CA
2
+AP
2
+2CA.AP−DB
2
−BP
2
−2BD.BP
CD
2
+DB
2
=CA
2
+AP
2
−BP
2
+2CA.AP−2DB.BP
CB
2
=CA
2
−(BP
2
−AP
2
)+2CA.AP−2DB.BP
CB
2
=CA
2
−AB
2
+2CA.AP−2DB.BP
CB
2
−CA
2
+AB
2
=2CA.AP−2DB.BP
AB
2
+AB
2
=2CA.AP−2DB.BP
2AB
2
=2[CA.AP−DB.BP]
AB
2
=(PC−AP)AP−(DP−BP)BP
AB
2
=AP×PC−AP
2
−DP×BP+BP
2
AB
2
=AP.PC−DP.BP+AB
2
=AP×PC−DP×BP
⇒AP×PC=DP×BP
Hence proved.