Question

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

(i) b² = p² + ax + (a²/4)

(ii) c² = p² – ax + (a²/4)

(iii) b² + c² = 2p² + (a²/2)

Answer 1

From the figure, D is the mid point of BC.

We have ∠AED = 90°

∴ ∠ADE < 90° and ∠ADC > 90°

i.e. ∠ADE is acute and ∠ADC is obtuse,

(i) In ΔADC, ∠ADC is obtuse angle.

AC² = AD² + DC² + 2DC x DE

⇒ AC² = AD² + (1/2)BC² + 2.(1/2)BC . DE

⇒ AC² = AD² + (1/4)BC² + BC.DE

⇒ AC² = AD² + BC.DE + (1/4)BC²

⇒ b² = p² + ax + (1/4)a²

Hence proved.

(ii) In ΔABD, ∠ADE is an acute angle.

AB² = AD² + BD² – 2BD.DE

⇒ AB² = AD² + ((1/2)(BC))² – 2 x (1/2)BC.DE

⇒ AB² = AD² + (1/4)BC² – BC.DE

⇒ AB² = AD² – BC.DE + (1/4)BC²

⇒ c² = p² – ax + (1/4)a²

Hence proved.

(iii) From (i) and (ii) we get .

AB² + AC² = 2AD² + (1/2)BC²

i.e. c² + b² = 2p² + (a²/2)

Hence it is proved.