+77 In the diagram below, quadrilateral $ABDE$ is a parallelogram, and $BC = BD$. If $\angle CBD = 28^\circ$, then find $\angle BAE$, in degrees.
+9481 I found this diagram from here: https://www.mathisfunforum.com/viewtopic.php?id=22434
There are 180° in every triangle. So..
| m∠BDC | + | m∠BCD | + | m∠CBD | = | 180° | ||
| The problem says that m∠CBD = 28° | ||||||||
| m∠BDC | + | m∠BCD | + | 28° | = | 180° | ||
| Since BD = BC, m∠BCD = m∠BDC | ||||||||
| m∠BDC | + | m∠BDC | + | 28° | = | 180° | ||
| 2(m∠BDC) | + | 28° | = | 180° | ||||
| Subtract 28° from both sides. | ||||||||
| 2(m∠BDC) | = | 152° | ||||||
| Divide both sides of the equation by 2 . | |||||||
| m∠BDC | = | 76° | ||||||
And since m∠BDC and m∠BDE form a straight line, they sum to 180° .
76° + m∠BDE = 180°
Subtract 76° from both sides.
m∠BDE = 104°
In a parallelogram, opposite angles are congruent. Imagine sliding up ED onto AB and sliding over BD onto AE. Then you can see that they are vertical angles, so they have the same measure.
m∠BDE = m∠BAE = 104°
