Q32 of 40 Page 1

Prove that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let us consider a right triangle ABC right-angled at B.

To Prove: AB2 + BC2 = AC2


Construction: Draw BD ⊥ AC



In ΔADB and ΔABC, we have


∠BAD = ∠BAC [Common]


∠ABC = ∠ADB [Both 90°]


⇒ ΔADB ∼ ΔABC [By AA similarity]


Sides of similar triangles are proportional, therefore



⇒ AD.AC = AB2 [1]


In ΔBDC and ΔABC, we have


∠BCD = ∠BCA [Common]


∠ABC = ∠BDC [Both 90°]


⇒ ΔBDC ∼ ΔABC [By AA similarity]


Sides of similar triangles are proportional, therefore



⇒ CD.AC = BC2 [2]


Adding [1] and [2], we get


AD.AC + CD.AC = AB2 + BC2


⇒ AC(AD + CD) = AB2 + BC2


⇒ AC.AC = AB2 + BC2


⇒ AC2 = AB2 + BC2


Hence, Proved!


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