In the figure, we have $\widehat{B}=50^{\circ}$, with $A D$ and $C D$ being the bisectors of angles $\widehat{A}$ and $\widehat{C}$, respectively. What is the measure of angle $A \widehat{D} C$?

(a) $90^{\circ}$

(b) $100^{\circ}$

(c) $115^{\circ}$

(d) $122.5^{\circ}$

(e) $125^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-002.jpg?height=502&width=371&top_left_y=800&top_left_x=1462)
The correct option is (c).

In this question, we will use an important theorem of Plane Geometry, as follows.

Theorem: The sum of the interior angles of a triangle is always $180^{\circ}$.

By the theorem, we have $\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}$ and, since $\widehat{B}=50^{\circ}$, it follows that $\widehat{A}+50^{\circ}+\widehat{C}=180^{\circ}$, that is, $\widehat{A}+\widehat{C}=130^{\circ}$. Since $A D$ and $C D$ are the bisectors of the angles $\widehat{A}$ and $\widehat{C}$, respectively, the theorem applied to the triangle $\triangle A D C$ gives the relation

$$
\frac{1}{2} \widehat{A}+\frac{1}{2} \widehat{C}+A \widehat{D} C=180^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-036.jpg?height=557&width=406&top_left_y=1292&top_left_x=1476)

But $\frac{1}{2} \widehat{A}+\frac{1}{2} \widehat{C}=\frac{1}{2}(\widehat{A}+\widehat{C})=\frac{1}{2} 130^{\circ}=65^{\circ}$, therefore, from the above equality, it follows that $A \widehat{D} C=180^{\circ}-65^{\circ}=115^{\circ}$.