The quadrilateral ABCD is a rhombus with BD = BC. On the side

1 answer:

5 0

The halves of the rhombus separated by line BD are equilateral triangles.
∠BAD = 60°
so
∠ADM is 0.80*60° = 48°.

∠BMD is an exterior angle to ΔAMD, so is equal to the sum
∠BMD = ∠BAD + ∠ADM
= 60° +48° = 108°

∠BMD = 108°

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