Question

Let $ABC$ be an isosceles triangle with $AB=AC$. Let be its circumcircle and let $O$ be the centre of Let $CO$ meet in $D$. Draw a line parallel to $AC$ through $D$. Let it intersect $AB$ at $E$. Suppose $AE:EB=2:1$ then $ABC$ is an equilateral triangle.

• A. True
• B. False

Answer 1

The correct option is A True

Extend DE to meet BC at F.

Join BD and DA. Since CD is a diameter,

we see that ∠DBC = 90◦ .

Since DF || AC,

it follows that $\Delta$EBF ∼ $\Delta$ABC.

Hence EB = EF.

Now $\angle EBF=\angle EFB=90◦-\angle EDB.$$

But $\angle EBF+\angle EBD=90◦$ .

Hence we obtain ∠EBD = ∠EDB, which gives EB = ED.

Since AE : EB = 2 : 1 and EB = ED,

we obtain AE = 2ED. Hence ∠DAB = 30◦

. This implies ∠DCB = 30◦ and hence ∠BDC = 60◦ .

But then ∠BAC = ∠BDC = 60◦ and

hence $\Delta$ABC is equilateral