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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 $AB^{2}+BC^{2}=AC^{2}$
Updated on: 10-Oct-2022
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