Question

In the quadrilateral ABCD, ∠B = ∠D = 90°. Prove that 2AC² – BC² = AB² + AD² + DC².

[Hint: LHS = (AC²) + (AC² – BC²) = (AD² + CD²) + (AB²) = RHS]

Answer 1

Given: In the quadrilateral ABCD, ∠B = ∠D = 90°.

To prove: 2AC² – BC² = AB² + AD² + DC²

Proof: Let’s place the quadrilateral on the coordinate plane:

Place B at origin (0, 0).

Let A = (0, a) so that AB = a

Let C = (b, 0) so that BC = b

Since ∠D = 90° and assuming D lies such that ∠D is at right angle between AD and DC, let’s assign coordinates to D = (b, d)

Then:

1. Use the distance formula:

AB² = (0 – 0)² + (a – 0)² = a²

BC² = (b – 0)² + (0 – 0)² = b²

AD² = (b – 0)² + (d – a)² = b² + (d – a)²

DC² = (b – b)² + (d – 0)² = d²

AC² = (b – 0)² + (0 – a)² = b² + a²

2. Left-Hand Side (LHS):

2AC² – BC² = 2(b² + a²) – b² = b² + 2a²

3. Right-Hand Side (RHS):

AB² + AD² + DC² = a² + [b² + (d – a)²] + d²

Let’s expand (d – a)²:

(d – a)² = d² – 2ad + a²

Now plug in:

a² + b² + d² – 2ad + a² + d²

= a² + a² + b² + d² + d² – 2ad

= 2a² + b² + 2d² − 2ad

But this is not matching LHS yet.

Let’s use pure geometry no coordinates.

We will apply Pythagoras Theorem in right triangles:

In triangle ABC:

AB² + BC² = AC²

⇒ AB² = AC² – BC²   ...(1)

In triangle ADC:

AD² + DC² = AC²

⇒ AD² + DC² = AC²   ...(2)

Add equations (1) and (2):

AB² + AD² + DC² = AC² – BC² + AC² = 2AC² – BC²

Thus, 2AC² – BC² = AB² + AD² + DC²

Hence proved.