Question

Prove that in a right angled triangle square of the hypotenuse is equal to sum of the squares

of other two sides.❤❤❤

Answer 1

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Answer 2

In ΔABC, m∠ABC=90°

Construction: BD is a perpendicular on side AC

To Prove: (AC)²=(AB)²+(BC)²

Proof:

In △ABC,

m∠ABC=90° (Given)

seg BD is perpendicular to hypotenuse AC (Construction)

Therefore, △ADB∼△ABC∼△BDC (Similarity of right-angled triangle)

△ABC∼△ADB

(AB/AC)=(AD/AB) (congruent sides of similar triangles)

AB2=AD×AC (1)

△BDC∼△ABC

CD/BC=BC/AC (congruent sides of similar triangles)

BC2=CD×AC (2)

Adding the equations (1) and (2),

AB2+BC2=AD×AC+CD×AC

AB2+BC2=AC(AD+CD)

Since, AD + CD = AC

Therefore, AC2=AB2+BC2