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

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#### Solution

**Given:** A right ΔABC right angled at B

**To prove :** AC^{2} = AB^{2} + BC^{2}

**Construction:** Draw AD ⊥ AC

**Proof:** ΔADB and ΔABC

∠ADB = ∠ABC = 90°

∠BAD = ∠BAC (common)

∴ ΔADB ∼ ΔABC (by AA similarly criterion)

`=> (AD)/(AB) = (AB)/(AC)`

⇒ AD × AC = AB^{2} ...... (1)

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA (common)

∴ ΔBDC ∼ ΔABC (by AA similarly criterion)

`=> (CD)/(BC) = (BC)/(AC)`

⇒ CD × AC = BC^{2} ........ (2)

Adding (1) and (2) we get

AB^{2} + BC^{2} = AD × AC + CD × AC

= AC (AD + CD)

= AC × AC = AC^{2}

∴ AC^{2} = AB^{2} + BC^{2}

Hence Proved.

Concept: Similarity of Triangles

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