In the triangle $ABC$ below, there is a point $D$ on its side $AB$, such that $AB=CD$, $\widehat{ABC}=100^{\circ}$, and $\widehat{DCB}=40^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-40.jpg?height=455&width=894&top_left_y=478&top_left_x=460)

a) What is the measure of the angle $\widehat{BDC}$?

b) What is the measure of the angle $\widehat{ACD}$?

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Solution

a) Analyzing the sum of the internal angles of the triangle $\triangle B D C$, we have $100^{\circ}+40^{\circ}+\angle B D C=180^{\circ}$, that is, $\angle B D C=40^{\circ}$.

b) Due to the previous item, $\triangle B D C$ is isosceles with base $C D$ and thus $B C=B D$. Construct the auxiliary triangle $\triangle B C E$, equilateral, and denote $\angle A C D=\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-40.jpg?height=442&width=849&top_left_y=1624&top_left_x=506)

By construction, $\angle D C E=100^{\circ}$ and $C E=B C$. From $A B=D C$ and $\angle A B C=\angle D C E$, we can write $\triangle A B C \equiv \triangle D C E$, by the $L A L$ case. Thus, we arrive at

$$
\angle A C B=\angle D E C=40^{\circ}+\alpha
$$

Since $\triangle B C E$ is equilateral ($B C=C E=E B$); and $\triangle B C D$ is isosceles ($B C=B D$), we conclude that $\triangle B D E$ is isosceles with $B E=B D$ and base $D E$. Therefore, its base angles are equal to $10^{\circ}$. Finally, since $\angle B E C=60^{\circ}=10^{\circ}+40^{\circ}+\alpha$, we conclude that $\alpha=10^{\circ}$.