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# In the given figure, ABD is a triangle right angled at A and $AC \perp BD$. Show that $AD^2= BD.CD$

"

Given:

$\triangle ABD$ is a right triangle right-angled at A and $AC \perp BD$.

To do:

We have to show that $AD^2=BD.CD$.

Solution:

In $\triangle DCA$ and $\triangle DAB$,

$\angle DCA=\angle DAB=90^o$

$\angle CDA=\angle BDA$ (Common angle)

Therefore,

$\triangle DCA \sim\ \triangle DAB$ (By AA similarity)

This implies,

$\frac{DC}{DA}=\frac{DA}{DB}$ (Corresponding parts of similar triangles are proportional)

$DA.DA=DB.DC$ (By cross multiplication)

$AD^2=BD.CD$

Hence proved.

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