Question

In triangle ABD, C is a point on BD such that BC:CD=1:2 and triangle ABC is an equilateral triangle then prove AD²=7AC²​

Answer 1

Answer:

Here ABC is an equilateral triangle, where BD = 3BC

By using Pythagoras we get the following:

As ABC is equilateral,

AB = BC = AC

BD = 3BC = 3AC

Let CH and AG be at right angles to AB

As ABC is equilateral and triangles HBC = ECF = GDF (should be apparent from the figure, so I won’t use space to prove it here)

AH = CE = GD = ½AB = ½AC

According to Pythagoras:

AE² = AC² – CE² = AC² – (½AC)² = ¾AC²

Therefore:

AD² = AG² + GD²

= (3AE)² + (½AC)²

=9 * ¾AC² + ¼AC²

= (27/4 + 1/4)AC²

= 28/4 AC²

= 7AC²

i.e. AD² = 7AC²

Which we were asked to prove.