Question

In rectangle ABCD, the diagonals intersect at E. If m angle∠AEB=  3x and m angle∠DEC= x+80, find m angle∠AEB and m angle∠EBA.

Answer 1

Since the angles∠ AEB and ∠DEC are vertically opposite angles, they are congruent, so we have:

`\begin{gathered} 3x=x+80 \\ 2x=80 \\ x=40 \end{gathered}`

So the measure of angle ∠AEB is:

`\begin{gathered} \angle\text{AEB}=3x \\ \angle\text{AEB}=3\cdot40=120\degree \end{gathered}`

The diagonals of a rectangle are congruent and intersect in their middle point, so the segment AE is congruent to the segment EB, therefore the triangle AEB is isosceles, so the angle ∠BAE is congruent to ∠EBA.

The sum of the internal angles of a triangle is 180°, so in triangle AEB we have:

`\begin{gathered} \angle\text{BAE}+\angle\text{EBA}+\angle\text{AEB}=180\degree \\ \angle\text{EBA}+\angle\text{EBA}+120=180 \\ 2\angle\text{EBA}=60 \\ \angle\text{EBA}=30\degree \end{gathered}`