Exercise 5. Let $A B C D$ be a convex quadrilateral, with $\widehat{A B C}=90^{\circ}, \widehat{B A D}=\widehat{A D C}=80^{\circ}$. Let $M$ and $N$ be points on $[A D]$ and $[B C]$ such that $\widehat{C D N}=\widehat{A B M}=20^{\circ}$. Suppose finally $M D=A B$. What is $\widehat{M N B}$?
Solution to exercise 5 We find many isosceles triangles in this figure!

![](https://cdn.mathpix.com/cropped/2024_04_29_d6d3fe511dcea719fb79g-3.jpg?height=615&width=504&top_left_y=487&top_left_x=802)

Let's prove all of this. The sum of the angles in triangle MAB is $180^{\circ}$, so $\widehat{A M B}=180^{\circ}-20^{\circ}-80^{\circ}=80^{\circ}$, hence $\mathrm{MAB}$ is isosceles at $B$. Therefore, MDB is isosceles at $\mathrm{M}$, and $\widehat{\mathrm{DMB}}=180^{\circ}-\widehat{\mathrm{AMB}}=100^{\circ}$.

Let's also calculate $\widehat{\mathrm{DNB}}=\widehat{\mathrm{NDC}}+\widehat{\mathrm{DCN}}=20^{\circ}+\left(360^{\circ}-80^{\circ}-80^{\circ}-90^{\circ}\right)=130^{\circ}$.

Now let $\mathcal{C}$ be the circle centered at $M$ with radius $M D=M B$. Since

$$
\widehat{\mathrm{DMB}}=100^{\circ}=2 \times 50^{\circ}=2 \times\left(180^{\circ}-130^{\circ}\right)=2 \times\left(180^{\circ}-\widehat{\mathrm{DNB}}\right)
$$

the inscribed angle theorem tells us that $\mathrm{N}$ lies on the circle $\mathrm{C}$.

Therefore, $M N=M B$, so $M N B$ is isosceles at $M$ and $\widehat{M N B}=\widehat{M B N}=70^{\circ}$.