In the given figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB= c, ED = x, AD =p and AE = h, prove that

(i) b² = p² + ax + a²/4

(ii)(b²+c²)=2p² + 1/2 a²

(iii) (b² —c²) = 2ax

Given: D is the mid-point of side BC and AE BC

and BC = a, AC = b, AB= c, ED = x, AD =p and AE = h

To Prove: (i)

or

Proof: In right triangle ∆AEC, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AE)² + (EC)² = (AC)²

⇒ (AE)² + (ED + DC)² = (AC)²

⇒ (AE)² + (ED)² + (DC)² + 2 (ED)(DC) = (AC)²

[∵ (a + b)² = a² + b² +2ab]

⇒ (AE² + ED²) + (DC)² + 2 (ED)(DC) = (AC)²

⇒ (AD)² + (DC)² + 2 (ED)(DC) = (AC)²

[∵ In right angled ∆AED, AE² + ED² =AD²]

[∵ D is the midpoint of BC, so 2DC = BC]

…(i)

To Prove: (ii)

or

Proof: In right triangle ∆AEB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AE)² + (BE)² = (AB)²

⇒ (AE)² + (BD – ED)² = (AB)²

⇒ (AE)² + (ED)² + (BD)² – 2 (ED)(BD) = (AB)²

[∵ (a – b)² = a² + b² – 2ab]

⇒ (AE² + ED²) + (BD)² – 2 (ED)(BD) = (AB)²

⇒ (AD)² + (BD)² – 2 (ED)(BD) = (AB)²

[∵ In right angled ∆AED, AE² + ED² =AD²]

[∵ D is the midpoint of BC, so 2DC = BC]

…(ii)

On adding eq. (i) and (ii), we get

To Prove: (iii) (b² —c²) = 2ax

or (AC)² – (AB)² = 2 (BC)(ED)

Proof: Subtracting Eq. (ii) from (i), we get

⇒ (AC)² – (AB)² = 2 (BC)(ED)

Hence Proved