Question

Prove that, in a right angled triangle, the square of the hypotenuse is

equal to the sum of the squares of remaining two sides.

Answer 1

Given:

A right angled ∆ABC, right angled at B

To Prove- AC²=AB²+BC²

Construction: draw perpendicular BD onto the side AC .

Proof:

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

△ADB∼△ABC. (by AA similarity)

Therefore, AD/ AB=AB/AC

(In similar Triangles corresponding sides are proportional)

AB²=AD×AC……..(1)

Also, △BDC∼△ABC

Therefore, CD/BC=BC/AC

(in similar Triangles corresponding sides are proportional)

Or, BC²=CD×AC……..(2)

Adding the equations (1) and (2) we get,

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)

( From the figure AD + CD = AC)

AB²+BC²=AC . AC

Therefore, AC²=AB²+BC²

This theroem is known as Pythagoras theroem...

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Hope this will help you.....

Answer 2

given in triangle abc angle abc is 90

draw perpendicular seg bd on side ac

abc is similar to adb and corresponding side is ab square =ad multiply ac similarly prove bc square = dc multiply ac

add 1and2

you will get answer