Question

Prove that in right angled triangle square of the hypotenuse is equal to sum of the square of other two sides

Answer 1

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Answer 2

Given:-

∆ ABC is a right triangle right angled at B and BD perpendicular to AC .

To prove:-

(AB)² = (AB)² + (BC)²

Construction:-

BD perpendicular to AC was drawn.

Proof:-

∆ ABC ~ ∆ ADB (°.° if a perpendicular drawn from the right vertex of a Triangles then Triangles form either side to the perpendicular are similar to each other and similar to the whole Triangle)

$so \: \frac{AB}{AD} = \frac{AC}{AB}$

➬AB × AB = AC × AD.

➬(AB)² = AC × AD. _____ Eq 1

∆ CBA ~ ∆ CDB (°.° if a perpendicular drawn from the right vertex of a Triangles then Triangles form either side to the perpendicular are similar to each other and similar to the whole Triangle)

$so, \: \frac{CB}{CD} = \frac{CA}{CB}$

➬CB × CB = CA × CD

➬(CB)² = CA. CD _____ Eq 2.

adding equation 1 and 2.

AB² + CB² = ( AC × AD ) + ( AC × CD )

➬ AB² + CB² = AC( AD + CD )

➬AB² + CB² = AC × AC .

➬AB² + CB² = AC²

➬AC² = AB² + CB².

HENCE PROVED