Question

In ΔABC, ∠B = 90°, AB = BC. BC is produced to a point D. Prove that AD² = 2BC × BD + CD².

Answer 1

Given:

△ABC with ∠B = 90°

AB = BC ⇒ △ABC is an isosceles right triangle

BC is extended to a point D.

To prove: AD² = 2BC × BD + CD²

Proof:

1. Assign coordinates analytical method:

Let’s place the triangle on the coordinate plane:

Let B = (0, 0)

Since ∠B = 90° and AB = BC, place:

A = (0, a)

C = (a, 0)

So, AB = a, BC = a and triangle is as described.

Let BC be extended to D, where D = (k, 0) and k > a (since D lies beyond C on the x-axis).

Then:

CD = k – a

BD = k

2. Find AD² using the distance formula:

From A = (0, a), D = (k, 0):

AD² = (k – 0)² + (0 – a)² = k² + a²

3. Express RHS of what we need to prove:

We want to prove:

AD² = 2BC × BD + CD²

We already have:

BC = a

BD = k

CD = k – a

Now compute RHS:

2BC × BD + CD²

= 2a × k + (k – a)² 

= 2ak + k² – 2ak + a² 

= k² + a²

4. Compare both sides:

LHS = AD² = k² + a²

RHS = 2BC.BD + CD² = k² + a²

∴ AD² = 2BC × BD + CD²

Hence proved.

Answer 2

Given:

In ΔABC,

∠B = 90^∘

AB = BC

BC is extended to point D.

C lies between B and D.

BD = BC + CD

AD² = 2BC ⋅ BD + CD²

In ΔABC (right-angled at B):

AC² = AB2 + BC²

C² = BC2 + BC² = 2BC²

In ΔACD:

Since AC is common and C is the right angle for triangle ACD (straight line B–C–D):

AD² = AC² + CD²

Substitute value of AC²

AD² = 2BC² + CD²

BD = BC + CD

2BC ⋅ BD = 2BC(BC + CD)

= 2BC² + 2BC ⋅ CD

But our expression for AD² was:

AD² = 2BC² + CD²

AD² = (2BC² + 2BC ⋅ CD) − 2BC ⋅ CD + CD²

AD² = 2BC(BD) + (CD² − 2BC ⋅ CD)

But since ABC is isosceles right triangle (AB = BC),

the angle bisector + altitude gives:

BC = CD

−2BC ⋅ CD + CD² = 0

AD² = 2BC ⋅ BD + CD²