In the adjoining figure, $A C ⊥ B D$ and $C E ⊥ A D$ . Prove that $A C^{2} = D A \times A E$ .

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Solution

To prove:- ${A C}^{2} = A D \times A E$
Proof:-
From the given fig., in $△ A C D$ and $△ A E C$ ,
As we know that if a perpendicular is drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
$∴ △ A C D \sim △ A E C$
Also, if two triangles are similar, then the ratio of their corresponding sides are also equal.
Therefore,
$\frac{A C}{A E} = \frac{A D}{A C}$
$\Rightarrow {A C}^{2} = A D \times A E$
Hence proved.

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