$A B C$ is an isosceles triangle in which $\angle A=92^{\circ} . C B$ is extended to a point $D$. What is the size of $\angle A B D$ ?
(A) $88^{\circ}$
(B) $44^{\circ}$
(C) $92^{\circ}$
(D) $136^{\circ}$
(E) $158^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-100.jpg?height=223&width=477&top_left_y=1008&top_left_x=1301)
$A B C$ is an isosceles triangle in which $\angle A=92^{\circ} . C B$ is extended to a point $D$. What is the size of $\angle A B D$ ?
(A) $88^{\circ}$
(B) $44^{\circ}$
(C) $92^{\circ}$
(D) $136^{\circ}$
(E) $158^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-450.jpg?height=263&width=553&top_left_y=1018&top_left_x=1361)

## Solution

Since $\angle A=92^{\circ}$ then $\angle A B C=\angle A C B=\frac{180^{\circ}-92^{\circ}}{2}=44^{\circ}$.

Therefore, $\angle A B D=180^{\circ}-44^{\circ}=136^{\circ}$.

ANSWER: (D)