3
$\begingroup$

Points E and F lie on the sides BC and CD of rectangle ABCD, the AEF is an equilateral triangle. point M is the midpoint of the AF. Prove that the triangle BCM is equilateral.

$\endgroup$
1
  • $\begingroup$ I have a proof that shows that MC = MB meaning that triangle BCM is certainly isosceles. Would that help? $\endgroup$
    – imranfat
    Oct 16, 2013 at 15:25

1 Answer 1

3
$\begingroup$

Now join. See that $\angle EMF$ is $90^{\circ}$. [As $\triangle AEF$ is equilateral & $M$ is the midpoint]. $\angle ECF$ is $90^{\circ}$.

So quadrilateral $EMFC$ is cyclic. So $\angle MCB = \angle MFE = 60^{\circ}$. Similarly, quadrilateral $AMEB$ is cyclic. So, $\angle MBE = \angle MAE = 60^{\circ}$. So in triangle $MBC$, $\angle MBC = \angle MCB = 60^{\circ}$. Therefore, $MBC$ is equilateral.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .