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

Answers (1)

Given $\Delta ABC$ is right-angled at $B$

To prove $\rightarrow AC^{2}=AB^{2}+BC^{2}$

$\rightarrow$ Draw $BD\perp AC$

Proof $\rightarrow$ Theorem $\rightarrow$ If a perpendicular is drawn from the vertex of the right angle of a right-angled to the hypotenuse then $\Delta$ on both side of the perpendicular are similar to the whole triangle and to each other

$\Delta AOB\sim \Delta ABC$

Since, sides of similar $\Delta$ are in the same ratio, hence

$\Rightarrow \frac{AD}{AB}=\frac{AB}{AC}$

$\Rightarrow AD.AC=AB^{2}------(i)$

$\Delta BDC\sim \Delta ABC$

Similarly here also the similar sides will have the same ratio.

$\Rightarrow \frac{CD}{BC}=\frac{BC}{AC}$

$\Rightarrow CD.AC=BC^{2}--------(ii)$

Add eq (i) and (ii)

$AD.AC+CD.AC=AB^{2}+BC^{2}$

$AC(AD+CD)=AB^{2}+BC^{2}$

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

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

Hence proved

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Safeer PP

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