# Prove that in a right angle triangle, the square of the hypotenuse is equal the sum of squares of the other two sides.

Given: A right - angled triangle ABC in which $\angle B=90^{o}$

To do: To Prove : $( Hypotenuse)^{2}=( Base)^{2}+( Perpendicular)^{2}$

Or

In a $\vartriangle ABC:\ \ \ AC^{2}=AB^{2}+BC^{2}$

Solution:
Construction: From B draw $BD \perp AC$.

Proof:

In triangle $\vartriangle ADB$ and $\vartriangle ABC$,

We have

$\angle ADB=\angle ABC$                                    [Each equal to $90^{o}$]

And, $\angle A=\angle A$                                           [Common]

So, by AA - similarity criteria, we have

$\vartriangle ADB ~\vartriangle ABC$

$\frac{AD }{AB}=\frac{AB}{AC}$                   [In similar triangles corresponding sides are proportional]

$AB^{2}=AD\times AC$                             .......$( 1)$

In triangles $\vartriangle BDC$ and $\vartriangle ABC$, we have

$\angle CDB=\angle ABC$                                                         [Each equal to $90^{o}$]

And,  $\angle C=\angle C$                                                                    [Common]

So, by AA-similarity criteria, we have

$\vartriangle BDC~\vartriangle ABC$

$\Rightarrow \frac{DC}{BC}=\frac{BC}{AC}$              [In similar triangles corresponding sides are proportional]

$\Rightarrow BC^{2}=AC\times DC$                                   .....$( 2)$

Adding equation $( 1)$ and $( 2)$, we get

$AB^{2}+BC^{2}=AD\times DC+AC\times DC$

$\Rightarrow AB^{2}+BC^{2}=AC(AC+DC)$

$\Rightarrow AB^{2}+BC^{2}=AC×AC=AC^{2}$

$\Rightarrow AB^{2}+BC^{2}=AC^{2}$

Hence, Proved that $AB^{2}+BC^{2}=AC^{2}$

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Updated on: 10-Oct-2022

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