Question

In ΔABC, AD ⊥ BC. D divides BC in the ratio 2 : 3. Prove that 5AC² = 5AB² + BC².

[Hint: Let BD = 2x, AD = y and DC = 3x.]

Answer 1

Given: In ΔABC, AD ⊥ BC.

D divides BC in the ratio 2 : 3, so BD = 2x and DC = 3x.

Let AD be y.

To prove: 5AC² = 5AB² + BC²

Proof: Let’s apply the Pythagoras Theorem in the two right-angled triangles:

In ΔABD,

AB² = AD² + BD²

= y² + (2x)²

= y² + 4x²   ...(1)

In ΔADC,

AC² = AD² + DC²

= y² + (3x)²

= y² + 9x²   ...(2)

Multiply (1) by 5;

5AB² = 5(y² + 4x²)

= 5y² + 20x²   ...(3)

Now add BC² to both sides of (3).

Since BC = BD + DC

= 2x + 3x

= 5x

So, BC² = (5x)²

= 25x²   ...(4)

Add (3) and (4):

5AB² + BC² = 5y² + 20x² + 25x²

= 5y² + 45x²   ...(5)

Now multiply (2) by 5.

5AC² = 5(y² + 9x²)

= 5y² + 45x²  ...(6)

Compare (5) and (6):

5AC² = 5AB² + BC²

Hence Proved.