The apex of the parallelogram and the midpoints of its two opposite sides form an equilateral triangle.

The apex of the parallelogram and the midpoints of its two opposite sides form an equilateral triangle. Find the angles of a parallelogram

1. Let’s make a drawing.

2. Consider the triangle BEF.

By condition, this is an equilateral triangle. Means:

BE = BF = EF;

∠FBE = ∠BEF = ∠EFB = 180 °: 3 = 60 °.

3. Find ∠AFB.

BC || AD, BF – secant. Means,

∠AFB = ∠FBE = 60 °, as criss-crossing interior.

4. Consider a triangle ABF.

In it AF = BF, since AF is half of AD and BF is half of BC, also AD = BC. Hence AF = BF.

Hence, triangle ABF is isosceles.

In an isosceles triangle, the angles at the base are equal. I.e

∠BAF = ∠FBA.

Let’s find them.

∠BAF + ∠FBA + ∠AFB = 180 °;

∠BAF + ∠FBA + 60 ° = 180 °;

∠BAF + ∠FBA = 180 ° – 60 °;

∠BAF + ∠FBA = 120 °;

∠BAF = ∠FBA = 120 °: 2;

∠BAF = ∠FBA = 60 °.

5. Find all the angles of the parallelogram.

A parallelogram has equal opposite angles.

∠BAF = ∠BCD = 60 °.

∠ABC = ∠ADC = ∠FBA + ∠FBE;

∠ABC = ∠ADC = 60 ° + 60 °;

∠ABC = ∠ADC = 120 °.