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# Explain the concept of the Pythagorean Theorem.

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## Step 1: State the Pythagorean TheoremIn a right-angled triangle, the square of the length of the hypotenuse is the sum of the square of the base and the square of the perpendicular.Thus, ${\left(\mathrm{Hypotenuse}\right)}^{2}={\left(\mathrm{Base}\right)}^{2}+{\left(\mathrm{Perpendicular}\right)}^{2}$.Step 2: Prove the Pythagorean TheoremConsider the right-angled triangle, $∆ABC$, such that $\angle B={90}^{\circ }$.Now, a perpendicular is drawn from the point $B$ to the side $CA$.It is known from the theorem that, The perpendicular drawn from the vertex of the right angle of a right-angled triangle to the hypotenuse then both sides of the perpendicular are similar to the whole triangle and to each other.Therefore, $∆ABD~∆ABC$.Thus, according to the condition of similarity, $\frac{AD}{AB}=\frac{AB}{AC}$.$⇒A{B}^{2}=AD×AC.....................\left(1\right)$.Therefore, $∆DBC~∆ABC$.Thus, according to the condition of similarity, $\frac{CD}{BC}=\frac{BC}{AC}$.$⇒B{C}^{2}=CD×AC.....................\left(2\right)$.Step 3: Add equation $\left(1\right)$ and $\left(2\right)$.$A{B}^{2}+B{C}^{2}=\left(AD×AC\right)+\left(CD×AC\right)\phantom{\rule{0ex}{0ex}}⇒A{B}^{2}+B{C}^{2}=AC\left(AD+CD\right)$From the given figure it is clear that, $\left(AD+CD\right)=AC$.$\therefore A{B}^{2}+B{C}^{2}=AC×AC\phantom{\rule{0ex}{0ex}}⇒A{B}^{2}+B{C}^{2}=A{C}^{2}\phantom{\rule{0ex}{0ex}}⇒A{C}^{2}=A{B}^{2}+B{C}^{2}$Hence, it is proven that, ${\left(\mathrm{Hypotenuse}\right)}^{\mathbf{2}}\mathbf{=}{\left(\mathrm{Base}\right)}^{\mathbf{2}}\mathbf{+}{\left(\mathrm{Perpendicular}\right)}^{\mathbf{2}}$.

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