In Fig., D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC =

1.	In Fig., 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) \(b^2 = p^2+a+\frac{{a}^2}{4}\)(ii) \(c^2 = p^2 -{ax}+\frac{a^2}{4}\)(iii)\(v^2+c^2 =2p^2+\frac{{a}^2}{2}\)
Answer» We have D is the midpoint of BC (i) In ΔAEC AC² = AE² + EC² b²= AE² + (ED + DC)² b² = AD² + DC² + 2 x ED x DC (Given BC = 2CD) b² = p² + (a/2)² + 2(a/2)x b² = p² + a²/4 + ax b² = p² + ax + a²/4 ………….. (i) (ii) In ΔAEB AB² = AE² + BE² c²= AD² - ED² + (BD - ED)² c² = p² - ED² + BD²+ ED² - 2BD x ED c² = P²+ (a/2)² - 2(a/2)²x c²= p²- ax + a²/4 ……………….(ii) (iii) Adding equ. (i)and(ii) we get b² + c² = 2p² + a²/2

In Fig., 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) \(b^2 = p^2+a+\frac{{a}^2}{4}\)(ii) \(c^2 = p^2 -{ax}+\frac{a^2}{4}\)(iii)\(v^2+c^2 =2p^2+\frac{{a}^2}{2}\)

Answer»

We have

D is the midpoint of BC

(i) In ΔAEC

AC² = AE² + EC²

b²= AE² + (ED + DC)²

b² = AD² + DC² + 2 x ED x DC

(Given BC = 2CD)

b² = p² + (a/2)² + 2(a/2)x

b² = p² + a²/4 + ax

b² = p² + ax + a²/4 ………….. (i)

(ii) In ΔAEB

AB² = AE² + BE²

c²= AD² - ED² + (BD - ED)²

c² = p² - ED² + BD²+ ED² - 2BD x ED

c² = P²+ (a/2)² - 2(a/2)²x

c²= p²- ax + a²/4 ……………….(ii)

(iii) Adding equ. (i)and(ii) we get

b² + c² = 2p² + a²/2