Question

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

Answer 1

Answer:

HEY THERE

HERE'S UR ANSWER

Given: A right triangle ABC right angled at B.

To Prove: AC^2 = AB^2 + BC^2

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)

AB^2= AD × AC … (i)

Similarly, Δ BDC ~ Δ ABC

BC^2 = CD × AC … (ii)

Adding (i) and (ii)

AB^2+ BC^2= (AD × AC) + (CD × AC)

AB^2+ BC^2 = AC × (AD + CD)

AB^2 + BC^2 = AC^2

Hence Proved.

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Answer 2

Answer:

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