Exercise 2. Let $A B C$ be an isosceles triangle at $A$. Let $M$ be the midpoint of the segment $[B C]$. Let $D$ be the symmetric point of $M$ with respect to the segment $[A C]$. We denote $x$ as the angle $\widehat{B A C}$. Determine, in terms of $x$, the value of the angle $\widehat{M D C}$.
## Solution to Exercise 2

![](https://cdn.mathpix.com/cropped/2024_04_29_7c73d4f4cc1e1eb4e334g-04.jpg?height=668&width=763&top_left_y=417&top_left_x=652)

Solution $n^{\circ} 1$: Since points $M$ and $D$ are symmetric with respect to the line $(A C)$, we have $C M = C D$. Triangle $D C M$ is isosceles at point $C$, so $\widehat{M D C} = \widehat{D M C}$. The sum of the angles in triangle $M D C$ is $180^{\circ}$, thus

$$
180^{\circ} = \widehat{M D C} + \widehat{D M C} + \widehat{M C D} = 2 \widehat{M D C} + \widehat{M C D}
$$

Since points $M$ and $D$ are symmetric with respect to the line $(A C)$, we have $\widehat{A C M} = \widehat{A C D}$, so $\widehat{M C D} = 2 \widehat{A C M} = 2 \widehat{A C B}$. And since triangle $A B C$ is isosceles and the sum of its angles is $180^{\circ}$, we have $\widehat{M C D} = 2 \widehat{A B C} = 180^{\circ} - \widehat{B A C}$. Therefore,

$$
\widehat{M D C} = \frac{180^{\circ} - \widehat{M C D}}{2} = \frac{180^{\circ} - (180^{\circ} - \widehat{B A C})}{2} = \frac{x}{2}
$$

Solution $n^{\circ} 2$: Since points $M$ and $D$ are symmetric with respect to the line $(A C)$, the lines $(M D)$ and $(A C)$ are perpendicular, and we denote their intersection point as $E$. By symmetry, we have $\widehat{C A D} = \widehat{M A C}$, and since the median and the angle bisector from $A$ coincide in the isosceles triangle $A B C$ at $A$, we have $\widehat{C A D} = \frac{x}{2}$. $M$ is also the foot of the altitude from vertex $A$, so $\widehat{A M C} = 90^{\circ}$, and by symmetry $\widehat{A D C} = 90^{\circ}$. In the right triangle $A E D$ at $E$, we have $\widehat{A D E} = 90^{\circ} - \frac{x}{2}$, and thus $\widehat{M D C} = 90^{\circ} - \widehat{A D E} = 90^{\circ} - (90^{\circ} - \frac{x}{2}) = \frac{x}{2}$.

Solution $n^{\circ} 3$: Similarly as before, we show that $\widehat{A M C} = 90^{\circ}$, $\widehat{M A C} = x / 2$, and the lines $(A C)$ and $(M D)$ are perpendicular. By a similar reasoning to the solution 2 but in triangle $A M C$, we obtain that

$$
\widehat{C M D} = 90^{\circ} - \widehat{E M A} = \widehat{M A C} = \frac{x}{2}
$$

We then conclude by using that triangle $C M D$ is isosceles at $C$ and thus $\widehat{C D M} = \widehat{C M D} = x / 2$. Solution $n^{\circ} 4$: Similarly as in the previous solution, we establish that $\widehat{A D C} = 90^{\circ}$ and $\widehat{E A D} = \widehat{E A M} = \frac{x}{2}$. By symmetry, we have $A M = A D$. Thus, triangle $A M D$ is isosceles at $A$ and the sum of its angles is $180^{\circ}$, so

$$
180^{\circ} = \widehat{M A D} + \widehat{D M A} + \widehat{M D A} = x + 2 \widehat{A D M}
$$

Thus, $\widehat{A D M} = 90^{\circ} - \frac{x}{2}$. We can then conclude as in the previous solution that $\widehat{M D C} = \frac{x}{2}$. Solution $n^{\circ} 5$: Similarly as in solution $n^{\circ} 2$, we establish that $\widehat{M A C} = \frac{x}{2}$ and $\widehat{C M A} = 90^{\circ} = \widehat{A D C}$. Points $M$ and $D$ are therefore both on the circle with diameter $[A C]$. According to the inscribed angle theorem, we have $\widehat{M D C} = \widehat{M A C} = \frac{x}{2}$.