Question

In 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) `b^2 = p^2 + ax + a^2/4`

(ii) `c^2 = p^2 - ax + a^2/4`

(iii) `b^2 + c^2 = 2p^2 + a^2/2`

Answer 1

We have, D as the mid-point of BC

(i) AC² = AE² + EC²

b² = AE² + (ED + DC)²           [By pythagoras theorem]

b² = AD² + DC² + 2DC × ED

`b^2=p^2+(a/2)^2+2(a/2)xxx`      [BC = 2CD given]

`rArrb^2=p^2+a^2/4+ax`            .........(i)

(ii) In ΔAEB, by pythagoras theorem

AB² = AE² + BE²

⇒ c² = AD² − ED² + (BD − ED)²          [By pythagoras theorem]

⇒ c² = p² − ED² + BD² + ED² − 2BD × ED

`rArrc^2=p^2+(a/2)^2-2(a/2)xx x`

`rArrc^2=p^2+a^2/4-ax`                    .........(ii)

(iii) Add equations (i) and (ii)

`b^2+c^2=p^2+a^2/4+ax + p^2+a^2/4-ax`

`b^2+c^2=2p^2+(2a^2)/4`

`b^2+c^2=2p^2+a^2/2`