Answer :

Draw a BD

In ∆ ADB and ∆ ABC

BAD = BAC (same angle)

ADB = ABC (90⁰)

By the AA criterion for triangle, similarity which states that if two triangles have two pairs of congruent angles, then the triangles are similar.

∆ ADB ≈ ∆ ABC

AB2 = AC×AD …(1)

In ∆ BDC and ∆ ABC

DCB = ACB (same angle)

BDC = ABC (90⁰)

By AA theorem

∆ BDC ≈ ∆ ABC

BC2 = AC×CD …(2)

On adding eq(1) and eq(2)

AC×AD + AC×CD = AB2 + BC2

AC(AD + CD) = AB2 + BC2

AC×AC = AB2 + BC2

AC2 = AB2 + BC2

Hence proved.

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