$\triangle ABD$ is a right triangle right-angled at A and $AC \perp BD$. Show that
$AB^2=BC.BD$

Given:

$\triangle ABD$ is a right triangle right-angled at A and $AC \perp BD$.
To do:
We have to show that $AB^2=BC.BD$.

Solution:

In $\triangle ABD$ and $\triangle ABC$,

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

$\angle B=\angle B$   (Common angle)

Therefore,

$\triangle ADB \sim\ \triangle CAB$   (By AA similarity)

This implies,

$\frac{AB}{CB}=\frac{BD}{AB}$   (Corresponding parts of similar triangles are proportional)

$AB.AB=CB.BD$   (By cross multiplication)

$AB^2=BC.BD$

Hence proved.

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Updated on: 10-Oct-2022

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