Homework: Is the triangle an equilateral triangle?

Question

A triangle is drawn inside a circle with it's corners on the circle alongside the following conditions: $$AB = AC,$$ $$\angle ABM = \angle CBM$$ and $MB$ is the radius which equal $1$ length unit.

How to prove the triangle also is an equilateral triangle?

Answer 1

Thanks for the help everyone!

$△BMC => ∠MCB = ∠MBC$ because $MB = MC$ Same goes for $△AMB$ which is $≅$ to $△CMA$ because three equal sides. This means $∠ABM = ∠BCM = ∠CAM = ∠ACM = ∠CBM = ∠BAC$ (every angle is the same except the middle ones). Adding the angles sharing a side and you get three equal angels which mean it is an equilateral triangle.

Answer 2

You mean that you want to show the triangle is equilateral, not isosceles. In fact, this is true. Draw in $AM$ and $CM$ and consider all the angles. Remember that you know that $\angle ABC \cong \angle ACB$.

Answer 3

Hint: Draw in $AM$ and $CM$. You will get three isosceles triangles.

Answer 4

Since $AB=AC$, we obtain $\gamma=\beta$ and

since $\measuredangle ABM=90^{\circ}-\gamma$ and $\measuredangle MBC=90^{\circ}-\alpha$, we obtain $\gamma=\alpha$ and we are done!