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Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
Given: A right - angled triangle ABC in which $\angle B=90^{o}$
To do: To Prove : $(Hypotenuse)^{2}=(Base)^{2}+(Perpendicular)^{2}$
$AC^{2}=AB^{2}+BC^{2}$
Solution:
Construction: from B draw BD⟂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]
$AB^{2}=AD×AC .......( 1)$
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 criterian, 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
$AB^{2}+BC^{2}=AD×DC+AC×DC$
$\Rightarrow AB^{2}+BC^{2}=AC(AC+DC)$
$\Rightarrow AB^{2}+BC^{2}=AC×AC=AC^{2}$
$\Rightarrow AB^{2}+BC^{2}=AC^{2}$
Hence, Proved that $AB^{2}+BC^{2}=AC^{2}$
Updated on: 10-Oct-2022
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