Question

In the adjoining figure, AB > AC, BE = EC and ∠ADC = 90°.

Prove that:

1. AB² – AC² = 2BC.ED
2. AB² + AC² = 2(AE² + BE²)

Answer 1

Given: In the adjoining figure AB > AC, BE = EC and ∠ADC = 90°.

To Prove:

1. AB² – AC² = 2BC.ED
2. AB² + AC² = 2(AE² + BE²)

Proof [Step-wise]:

1. Coordinate the setup of a convenient frame.

Put BC on the x-axis and choose E as the origin.

Let BE = EC = a

So, B = (–a, 0)

E = (0, 0)

C = (a, 0)

Hence, BC = 2a.

Let D = (d, 0) for some d with –a < d < a.

Since ∠ADC = 90°, A lies on the line through D perpendicular to BC.

So, write A = (d, h) with h ≠ 0 take h > 0.

The given AB > AC implies AB² – AC² > 0 and as computed below forces d > 0; we keep d as the signed x-coordinate of D.

2. Express squared distances.

AB² = (x_A – x_B)² + (y_A – y_B)² 

= (d – (–a))² + h²

= (d + a)² + h²

AC² = (d – a)² + h²

AE² = (d – 0)² + h²

= d² + h²

BE² = (–a – 0)² + 0²

= a²

ED = Distance from E(0, 0) to D(d, 0) = |d|. 

With AB > AC we will have d > 0.

So, ED = d.

3. Prove (i): AB² – AC² = 2 × BC × ED.

Compute the difference:

AB² − AC²

= [(d + a)² + h²] – [(d – a)² + h²] 

= (d + a)² – (d – a)²

Expand or use difference of squares: 

(d + a)² – (d – a)² 

= [d² + 2ad + a²] – [d² – 2ad + a²] 

= 4ad

Since BC = 2a and ED = d because AB > AC gives d > 0.

2 × BC × ED

= 2 × (2a) × d

= 4ad

Hence, AB² – AC² = 4ad = 2 × BC × ED, as required.

4. Prove (ii): AB² + AC² = 2(AE² + BE²).

Sum the squares:

AB² + AC²

= [(d + a)² + h²] + [(d – a)² + h²] 

= (d + a)² + (d – a)² + 2h²

Compute (d + a)² + (d – a)² = 2(d² + a²).

So, AB² + AC² 

= 2(d² + a²) + 2h² 

= 2(d² + h² + a²)

But AE² + BE² = (d² + h²) + a².

So, 2(AE² + BE²) = 2(d² + h² + a²).

Therefore, AB² + AC² = 2(AE² + BE²), as required.