Subfield: Geometry


Question: Let $A B C D E$ be a convex pentagon such that $B C \| A E, A B=B C+A E$, and $\angle A B C=$ $\angle C D E$. Let $M$ be the midpoint of $C E$, and let $O$ be the circumcenter of triangle $B C D$. Given that $\angle D M O=90^{\circ}$, prove that $2 \angle B D A=\angle C D E$.

Solution:
Choose point $T$ on ray $A E$ such that $A T=A B$; then from $A E \| B C$ we have $\angle C B T=\angle A T B=\angle A B T$, so $B T$ is the bisector of $\angle A B C$. On the other hand, we have $E T=A T-A E=A B-A E=B C$, hence quadrilateral $B C T E$ is a parallelogram, and the midpoint $M$ of its diagonal $C E$ is also the midpoint of the other diagonal $B T$.

Next, let point $K$ be symmetrical to $D$ with respect to $M$. Then $O M$ is the perpendicular bisector of segment $D K$, and hence $O D=O K$, which means that point $K$ lies on the circumcircle of triangle $B C D$. Hence we have $\angle B D C=\angle B K C$. On the other hand, the angles $B K C$ and $T D E$ are symmetrical with respect to $M$, so $\angle T D E=\angle B K C=\angle B D C$.

Therefore, $\angle B D T=\angle B D E+\angle E D T=\angle B D E+\angle B D C=\angle C D E=\angle A B C=180^{\circ}-$ $\angle B A T$. This means that the points $A, B, D, T$ are concyclic, and hence $\angle A D B=\angle A T B=$ $\frac{1}{2} \angle A B C=\frac{1}{2} \angle C D E$, as desired.
<img_3951>
Let $\angle C B D=\alpha, \angle B D C=\beta, \angle A D E=\gamma$, and $\angle A B C=\angle C D E=2 \varphi$. Then we have $\angle A D B=2 \varphi-\beta-\gamma, \angle B C D=180^{\circ}-\alpha-\beta, \angle A E D=360^{\circ}-\angle B C D-\angle C D E=$ $180^{\circ}-2 \varphi+\alpha+\beta$, and finally $\angle D A E=180^{\circ}-\angle A D E-\angle A E D=2 \varphi-\alpha-\beta-\gamma$.
<img_4052>

Let $N$ be the midpoint of $C D$; then $\angle D N O=90^{\circ}=\angle D M O$, hence points $M, N$ lie on the circle with diameter $O D$. Now, if points $O$ and $M$ lie on the same side of $C D$, we have $\angle D M N=\angle D O N=\frac{1}{2} \angle D O C=\alpha ;$ in the other case, we have $\angle D M N=180^{\circ}-\angle D O N=\alpha ;$



so, in both cases $\angle D M N=\alpha$ (see Figures). Next, since $M N$ is a midline in triangle $C D E$, we have $\angle M D E=\angle D M N=\alpha$ and $\angle N D M=2 \varphi-\alpha$.

Now we apply the sine rule to the triangles $A B D, A D E$ (twice), $B C D$ and $M N D$ obtaining

$$
\begin{gathered}
\frac{A B}{A D}=\frac{\sin (2 \varphi-\beta-\gamma)}{\sin (2 \varphi-\alpha)}, \quad \frac{A E}{A D}=\frac{\sin \gamma}{\sin (2 \varphi-\alpha-\beta)}, \quad \frac{D E}{A D}=\frac{\sin (2 \varphi-\alpha-\beta-\gamma)}{\sin (2 \varphi-\alpha-\beta)} \\
\frac{B C}{C D}=\frac{\sin \beta}{\sin \alpha}, \quad \frac{C D}{D E}=\frac{C D / 2}{D E / 2}=\frac{N D}{N M}=\frac{\sin \alpha}{\sin (2 \varphi-\alpha)}
\end{gathered}
$$

which implies

$$
\frac{B C}{A D}=\frac{B C}{C D} \cdot \frac{C D}{D E} \cdot \frac{D E}{A D}=\frac{\sin \beta \cdot \sin (2 \varphi-\alpha-\beta-\gamma)}{\sin (2 \varphi-\alpha) \cdot \sin (2 \varphi-\alpha-\beta)}
$$

Hence, the condition $A B=A E+B C$, or equivalently $\frac{A B}{A D}=\frac{A E+B C}{A D}$, after multiplying by the common denominator rewrites as

$$
\begin{aligned}
& \sin (2 \varphi-\alpha-\beta) \cdot \sin (2 \varphi-\beta-\gamma)=\sin \gamma \cdot \sin (2 \varphi-\alpha)+\sin \beta \cdot \sin (2 \varphi-\alpha-\beta-\gamma) \\
& \Longleftrightarrow \cos (\gamma-\alpha)-\cos (4 \varphi-2 \beta-\alpha-\gamma)=\cos (2 \varphi-\alpha-2 \beta-\gamma)-\cos (2 \varphi+\gamma-\alpha) \\
& \Longleftrightarrow \cos (\gamma-\alpha)+\cos (2 \varphi+\gamma-\alpha)=\cos (2 \varphi-\alpha-2 \beta-\gamma)+\cos (4 \varphi-2 \beta-\alpha-\gamma) \\
& \Longleftrightarrow \cos \varphi \cdot \cos (\varphi+\gamma-\alpha)=\cos \varphi \cdot \cos (3 \varphi-2 \beta-\alpha-\gamma) \\
& \Longleftrightarrow \cos \varphi \cdot(\cos (\varphi+\gamma-\alpha)-\cos (3 \varphi-2 \beta-\alpha-\gamma))=0 \\
& \Longleftrightarrow \cos \varphi \cdot \sin (2 \varphi-\beta-\alpha) \cdot \sin (\varphi-\beta-\gamma)=0 .
\end{aligned}
$$

Since $2 \varphi-\beta-\alpha=180^{\circ}-\angle A E D<180^{\circ}$ and $\varphi=\frac{1}{2} \angle A B C<90^{\circ}$, it follows that $\varphi=\beta+\gamma$, hence $\angle B D A=2 \varphi-\beta-\gamma=\varphi=\frac{1}{2} \angle C D E$, as desired.