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The quadrilateral ABCD is a rhombus with BD = BC. On the side
AB
a point M is taken so that the measure of ∠ADM is 20% smaller than the measure of ∠BAD. Find the measure of ∠BMD.

Respuesta :

sqdancefan
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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