Question

In a figure, ABCDE is a regular pentagon. Bisector of angle A meets CD at M. Prove that angle AMC=90 degrees

Answer 1

See, I have circumscribed the regular Pentagon with a circle. Since, the 360° of the circle is divided equally by the pentagon to five parts, i.e., 72°, the angle COD =72°.

Now, I have calculated angle CAD=36°, which is half of angle COD, using the property of a circle.

Now, I have calculated angle ACD, using property of triangle, in ∆ACD.
36°+ angle ACD+ angle ADC= 180
Since, ∆ACD is an isosceles ∆,
36° +2× angle ACD= 180°
angle ACD = 72°

Now finally take a look at the ∆AMC.

Angle MAC = half of angle CAD= 18°
Now,
angle MAC+angle AMC+angle ACM=180°

18°+angle AMC + 72° = 180°

angle AMC = 90°
HENCE PROVED!