Two right triangles $A B C$ and $D B C$ are drawn on the same hypotenuse $B C$ and on the same sides of $B C$ . If $A C$ and $D B$ intersect at $P$ , prove that $A P \times P C = B P \times P D$ .

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Solution

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
$A P \times P C = B P \times P D$
as both are right angle triangles, let us use Pythagorean's theorem for the given triangles.
as per fig (a)
From $Δ C D P$ we get,
$C D^{2} = C P^{2} - D P^{2}$ ( $∵$ from fig (a) we get, $C P = C A + A P$ $D P = D B + B P$ )
$= (C A + A P)^{2} - (D B + B P)^{2}$
$[∵ (a + b)^{2} = a^{2} + 2 a b + b^{2}]$
$C D^{2} = C A^{2} + A P^{2} + 2 C A . A P - [D B^{2} + B P^{2} + 2 D B . B P]$
$C D^{2} = C A^{2} + A P^{2} + 2 C A . A P - D B^{2} - B P^{2} - 2 B D . B P$
$C D^{2} + D B^{2} = C A^{2} + A P^{2} - B P^{2} + 2 C A . A P - 2 D B . B P$
$C B^{2} = C A^{2} - (B P^{2} - A P^{2}) + 2 C A . A P - 2 D B . B P$
$C B^{2} = C A^{2} - A B^{2} + 2 C A . A P - 2 D B . B P$
$C B^{2} - C A^{2} + A B^{2} = 2 C A . A P - 2 D B . B P$
$A B^{2} + A B^{2} = 2 C A . A P - 2 D B . B P$
$2 A B^{2} = 2 [C A . A P - D B . B P]$
$A B^{2} = (P C - A P) A P - (D P - B P) B P$
$A B^{2} = A P \times P C - A P^{2} - D P \times B P + B P^{2}$
$A B^{2} = A P . P C - D P . B P + A B^{2}$
$= A P \times P C - D P \times B P$
$\Rightarrow A P \times P C = D P \times B P$
Hence proved.

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Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same sides of BC. If AC and DB intersect at P, prove that AP×PC=BP×PD.

Two right triangles $A B C$ and $D B C$ are drawn on the same hypotenuse $B C$ and on the same sides of $B C$ . If $A C$ and $D B$ intersect at $P$ , prove that $A P \times P C = B P \times P D$ .