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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