Question

plzz help me........

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Answer 1

Given: A right triangle ABC right angled at B.

To Prove: AC2 = AB2 + BC2

Construction: Draw BD ⊥ AC

Proof:

In Δ ADB and Δ ABC,

∠ ADB = ∠ ABC (each 90°)

∠ BAD = ∠ CAB (common)


Δ ADB ~ Δ ABC (By AA similarity criterion)

Now, AD/AB = AB/AC (corresponding sides are proportional)

AB2 = AD × AC … (i)

Similarly, Δ BDC ~ Δ ABC

BC2 = CD × AC … (ii)

Adding (i) and (ii)

AB2 + BC2 = (AD × AC) + (CD × AC)

AB2 + BC2 = AC × (AD + CD)

AB2 + BC2 = AC2

Hence Proved.

Answer 2

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