$△ B A C$ and $△$ BDC are two right-angled triangles with common hypotenuse BC. The sides $A C$ and $B D$ intersect at $^{'} P^{'}$
Prove that $A P . P C = D P . P B$

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Solution

In the given figure, $Δ B A C$ and $Δ C D B$ are right angled triangles.
In $Δ B A P$ and $Δ C D P$ ,
$∠ B A P = ∠ C D P = 90^{0}$
$∠ A P B = ∠ D P C$ (Vertically opposite angles)
Therefore, $∠ A B P = ∠ D C P$ (Remaining angles)

Thus $Δ B A P$ and $Δ C D P$ are similar by AA criterion.
$\Rightarrow \frac{A B}{C D} = \frac{A P}{P D} = \frac{P B}{P C}$

$\Rightarrow \frac{A P}{P D} = \frac{P B}{P C} \Rightarrow A P \times P C = D P \times P B$
Hence $A P \times P C = D P \times P B$ .

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△BAC and △BDC are two right-angled triangles with common hypotenuse BC. The sides AC and BD intersect at ′P′Prove that AP.PC=DP.PB

$△ B A C$ and $△$ BDC are two right-angled triangles with common hypotenuse BC. The sides $A C$ and $B D$ intersect at $^{'} P^{'}$
Prove that $A P . P C = D P . P B$