In the figure AB=AC, angle BAD=30°, and AE=AD. Then angle CDE equals:

1956 AMC 12 Problem 24

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24.

In the figure

AB=AC,

angle

BAD=30^\circ,

and

AE=AD.

Then angle

CDE

equals:

7\dfrac12^\circ

10^\circ

12\dfrac12^\circ

15^\circ

20^\circ

Let

\angle DAE=\alpha

and express the base angle at

C

using isosceles triangle

ABC

Use

AE=AD

to express

\angle ADE,

then split the straight angle at

D

Let

\angle DAE=\alpha.

Since

AB=AC

and

AE=AD,

\begin{aligned} \angle ACB&=75^\circ-\frac\alpha2,\\ \angle ADE&=90^\circ-\frac\alpha2. \end{aligned}

Triangle

ACD

gives

\angle ADC=180^\circ-\alpha-\angle ACB,

\angle ADC=105^\circ-\frac\alpha2.

Since

\angle ADC=\angle ADE+x,

\begin{aligned} x&=\left(105^\circ-\frac\alpha2\right) -\left(90^\circ-\frac\alpha2\right)\\ &=15^\circ. \end{aligned}

Thus, the correct answer is D.