Question

in equilateral triangle abc, BC perpendicular AD such that d is a point on BC then show that bc square = 4( AC square minus AD square)

Answer 1

Here, triangle ABC is an equilateral triangle,

That is, AB = BC= CA

And,AD⊥BC

Thus, by the property of equilateral triangle,

AD must be the bisector of side BC.

That is, BD = DC = {1}{2} BCBD=DC=21​BC

By the Pythagoras theorem,

AD^2 = AB^2 -BD^2AD2=AB2−BD2 ------(1)

And,  AD^2 = AC^2 -CD^2AD2=AC2−CD2 -------(2)

By adding equation (1) and (2),

We get,

2 AD^2 = AB^2 + AC^2 - (BD^2 + CD^2)2AD2=AB2+AC2−(BD2+CD2)

2 AD^2 =AC^2 + AC^2 - ((\frac{1}{2} BC)^2 + (\frac{1}{2} BC)^2)2AD2=AC2+AC2−((21​BC)2+(21​BC)2)

2 AD^2 = 2 AC^2 - (\frac{1}{4} BC^2 + \frac{1}{4} BC^2)2AD2=2AC2−(41​BC2+41​BC2)

2 AD^2 = 2 AC^2 - (\frac{1}{4} BC^2 + \frac{1}{4} BC^2)2AD2=2AC2−(41​BC2+41​BC2)

2 AD^2 = 2 AC^2 - \frac{1}{2} BC^22AD2=2AC2−21​BC2

2 AD^2 - 2 AC^2 = - \frac{1}{2} BC^22AD2−2AC2=−21​BC2

4 AD^2 - 4 AC^2 = - BC^24AD2−4AC2=−BC2

4 AD^2 - 4 AC^2 = - BC^24AD2−4AC2=−BC2

4 ( AC^2 - AD^2)= BC^24(AC2−AD2)=BC2

Hence proved.

hope this answer will help you.