Question

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Answer 1

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Answer 2

$\huge\underline\mathcal\red{Solution}$

Given :

In triangle ABC ...

<B = 90°

To prove:

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

Construction :

Draw BD _|_ AC

Proof :

Since in triangle ABC , <B= 90° and BD _|_ AC .

So , triangle ADB ~Triangle ABC

[ If a _|_ is drawn from the vertex of the right angle of right triangle to the hypotenuse then triangles on both side of the _|_ are similar to the whole triangle and to each other .]

and triangle BDC ~ Triangle ABC

Now , triangle ADB ~Triangle ABC

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

$=> {AB}^{2} = AC . AD ......(i)$

Again , Triangle BDC ~ triangle ABC

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

$=> {BC}^{2} = AC .CD ......(ii)$

Adding eq. (i) and (ii) , we get ,

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

= AC (AD + CD )

= AC . AC

= {AC }^{2}

Hence ,

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