Question

A point D is on the side BC of an equilateral triangle ABC such that \[DC = \frac{1}{4}BC\]. Prove that AD² = 13CD².

Answer 1

We are given ABC is an equilateral triangle with `CD = 1/4 BC`

We have to prove `AD^2 = 13DC^2`

Draw `AE  ⊥  BC`

In ΔAEB and ΔAED we have `AB = AC`

`∠AEB = ∠AEC = 90^circ`

`AE = AE`

So by right side criterion of similarity we have

Thus we have

`DC =1/4BC and BE =EC = 1/2 BC`

Since `∠ C= 60^circ`, therefore

`AD^2 =AC^2+DC^2 - 2DC xxEC`

`AD^2 = AC^2+(1/4BC)^2-2xx1/4BCxx1/2BC`

`AD^2=AC^2+1/16BC^2-2xx1/4BCxx1/2BC`

`AD^2=AC^2+1/16BC^2-BC^2`

We know that AB = BC = AC

`AD^2=BC^2+1/16BC^2-BC^2`

`AD^2=(16BC^2+1BC^2-4BC^2)/16`

`AD^2=13/16BC^2`

We know that `DC = 1/4 BC`

`4DC=BC`

Substitute `4DC=BC` in `AD^2 =13/16BC^2 ` we get

`AD^2 = 13/16xx(4DC)^2`

`AD^2 = 13/16xx16DC^2`

`AD^2 = 13/16xx16xxDC^2`

`AD^2=13DC^2`

Hence we have proved that `AD^2 = 13 DC^2`