Question

prove that in a right triangel , the square of the hypotenuse is equal to the sum of the square of the other two sides ​

Answer 1

$\huge\boxed{Answer}$

Given:- A right triangle ABC right angled at B.

To prove:- AC2=AB2+BC2

Construction:- Draw BD⊥AC

Proof:-

In △ABC and △ABD

∠ABC=∠ADB(Each 90°)

∠A=∠A(Common)

∴△ABC∼△ABD(By AA)

ACAB=ABAD(∵Sides of similar triangles are proportional)

⇒AB2=AD⋅AC.....(1)

Similarly, in △ABC and △BCD

∠ABC=∠BDC(Each 90°)

∠C=∠C(Common)

∴△ABC∼△BCD(By AA)

∴BCDC=ACBC

⇒BC2=DC⋅AC.....(2)

Adding equation (1)&(2), we have

AB2+BC2=AD⋅AC+DC.AC

⇒AB2+BC2

$\huge\mathfrak\red{itz\:jyotsana☺}$