Question

ABCD is a square and triangle EDC is an equilateral triangle . prove that AE=BE and angle DAE=15

Answer 1

Answer on Question #39637 – Math – Geometry

ABCD is a square and triangle EDC is an equilateral triangle. Prove that AE=BE and angle DAE=15

Solution:

As ABCD is a square, we have $\angle ADC = \angle BCD = 90{}^\circ$. As DEC is an equilateral triangle, then $\angle EDC = \angle ECD = 60{}^\circ$, and $\angle EDA = \angle EDC + \angle ADC = 150{}^\circ$, $\angle ECB = \angle ECD + \angle BCD = 150{}^\circ$.

In triangles ADE and BCE, AD = BC, $\angle EDA = \angle ECB$. As sides of equilateral triangle are equal, then ED = EC. Therefore $\Delta ADE = \Delta BCE$

Then AE = BE.

Since ABCD is a square, then AB = BC = CD = AD, (1)

since CDE is an equilateral triangle, then CD = DE = EC. (2)

From (1) and (2), we have

AB = BC = AD = CD = DE = EC. (3)

In triangle DAE, by (3)

AD = DE, then, as angles opposite to equal sides are equal, $\angle DEA = \angle DAE$.

In triangle DAE,

$\angle ADE + \angle DEA + \angle DAE = 180{}^\circ$

$150{}^\circ + 2\angle DEA = 180{}^\circ$

We get, $\angle DEA = \angle DAE = 15{}^\circ$

QED