In the diagram, $\angle A B D=\angle B D C$ and $\angle D A B=80^{\circ}$. Also, $A B=A D$ and $D B=D C$. The measure of $\angle B C D$ is
(A) $65^{\circ}$
(D) $60^{\circ}$
(B) $50^{\circ}$
(C) $80^{\circ}$
(E) $70^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-088.jpg?height=382&width=304&top_left_y=1034&top_left_x=1301)
In the diagram, $-A B D=-B D C$ and $-D A B=80$ r. Also, $A B=A D$ and $D B=D C$. The measure of $-B C D$ is
(A) $65 \mathrm{r}$
(B) $50 \mathrm{r}$
(C) $80 \mathrm{r}$
(D) $60 \mathrm{r}$
(E) $70 \mathrm{r}$

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-393.jpg?height=461&width=336&top_left_y=637&top_left_x=1447)

## Solution

Since $\mathrm{D} A B D$ is isosceles, then $-A B D=-A D B$.

Therefore,

$$
\begin{aligned}
80^{\circ}+-A B D+-A D B & =180^{\circ} \\
2 ¥-A B D & =100^{\circ} \quad(\text { since }-A B D=-A D B) \\
-A B D & =50^{\circ}
\end{aligned}
$$

Thus, $-B D C=50^{\circ}$ as well, since $-A B D=-B D C$. Since

$\mathrm{D} B D C$ is also isosceles, then if we repeat a similar calculation to above, we obtain that $-B C D=65^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-393.jpg?height=458&width=334&top_left_y=1132&top_left_x=1443)

ANSWER: (A)