Question

14. ABCDE is a regular pentagon. The bisector of
angle A of the pentagon meets the side CD in
point M. Show that ZAMC = 90°.
DM
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Answer 1

Answer:

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!

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