Exercise 5. Let ABCD 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_06_04_93e83a8cba80c1bd2fe8g-3.jpg?height=627&width=523&top_left_y=478&top_left_x=801)

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

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

Now let \(\mathcal{C}\) be the circle centered at \(M\) with radius \(MD = MB\). Since

\[
\widehat{\mathrm{DMB}} = 100^{\circ} = 2 \times 50^{\circ} = 2 \times (180^{\circ} - 130^{\circ}) = 2 \times (180^{\circ} - \widehat{\mathrm{DNB}})
\]

the theorem of the angle at the center tells us that \(N\) is on the circle \(\mathcal{C}\).

Therefore, \(MN = MB\), so \(MNB\) is isosceles at \(M\) and \(\widehat{MNB} = \widehat{MBN} = 70^{\circ}\).

## High School Statements