Using Pythagoras theorem determine the length of AD in terms of b and c shown in figure below.
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Given:

In $∆ABC$, $\angle A = 90^o$.

In $∆ABD$, $\angle ADB = 90^o$.

In $∆ACD$, $\angle ADC = 90^o$.

 To do:

We have to find the length of $AD$ in terms of $b$ and $c$.

Solution:

In $∆ABC$,

By Pythagoras theorem,

$BC^2 = AB^2 + AC^2$

$BC^2 = c^2 + b^2$

$BC = \sqrt{(c^2 + b^2)}$

In $∆ABD$ and $∆ABC$,

$\angle B = \angle B$   (Common)

$\angle ADB = \angle BAC = 90^o$         

Therefore,

$∆ABD ͏~ ∆CBA$   (By AA similarity)

This implies,

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

$\frac{c}{\sqrt{(c^2 + b^2)}} = \frac{AD}{b}$

$AD = \frac{bc}{\sqrt{(c^2 + b^2)}}$

The length of $AD$ in terms of $b$ and $c$ is $\frac{bc}{\sqrt{(b^2 + c^2)}}$.

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Updated on: 10-Oct-2022

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