Question

equilateral triangle ABC ad perpendicular BC such that D is a point on BC then show that BC square equal to4( AC square minus AD square)

Answer 1

Answer:

Here, triangle ABC is an equilateral triangle,

That is, AB = BC= CA

And, $AD\perp BC$

Thus, by the property of equilateral triangle,

AD must be the bisector of side BC.

That is, $BD = DC = \frac{1}{2} BC$

By the Pythagoras theorem,

$AD^2 = AB^2 -BD^2$ ------(1)

And,  $AD^2 = AC^2 -CD^2$ -------(2)

By adding equation (1) and (2),

We get,

$2 AD^2 = AB^2 + AC^2 - (BD^2 + CD^2)$

$2 AD^2 =AC^2 + AC^2 - ((\frac{1}{2} BC)^2 + (\frac{1}{2} BC)^2)$

$2 AD^2 = 2 AC^2 - (\frac{1}{4} BC^2 + \frac{1}{4} BC^2)$

$2 AD^2 = 2 AC^2 - (\frac{1}{4} BC^2 + \frac{1}{4} BC^2)$

$2 AD^2 = 2 AC^2 - \frac{1}{2} BC^2$

$2 AD^2 - 2 AC^2 = - \frac{1}{2} BC^2$

$4 AD^2 - 4 AC^2 = - BC^2$

$4 AD^2 - 4 AC^2 = - BC^2$

$4 ( AC^2 - AD^2)= BC^2$

Hence proved.

Answer 2

Solution:

An equilateral Δ AB C, in which , D is a point on side BC.

Such that, AD ⊥ BC

Also, AB=BC=CA

∵ CD = B D=$[\frac{BC}{2}]$→→In equilateral Δ, perpendicular acts as a median also.-----(1)

In Right Δ AD C

→AC²=AD²+DC²→→[By Pythagoras theorem]

→AC²-AD²= DC²

→AC²-AD²=$[\frac{BC}{2}]^2$→→using (1)

→BC²= 4 × [AC²-AD²]

Hence proved.