Solution to Problem 1315 of Gogeometry

 The following problem was proposed by my esteemed friend, Ángel Mejía. However, this is problem 1315 from Gogeometry. Here I give a trigonometric proof.

The figure shows a triangle $ABC$ with the inscribed circle $O$ ($D$, $E$, and $T$ are the tangency points). $OB$ cuts chord $DE$ and arc $DE$ at $M$ and $F$, respectively. $AF$ and $CF$ cut chord $DE$ at $G$ and $N$, respectively. Prove that $DG = MN$.

Proof. Let's denote $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$, $\angle{ACB}=\gamma$, $\angle{AFM}=\angle{AFO}=\delta$, and $\angle{DFA}=\epsilon$. Applying the sine law in triangles $GFM$ and $DFG$, we obtain

$$GM = FG \cdot \sin{\delta},\tag{1}$$

$$DG = \frac{FG \cdot \sin{\epsilon}}{\sin{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}.\tag{2}$$

Dividing $(1)$ by $(2)$ and then squaring, we get

$$\left(\frac{GM}{DG}\right)^2=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\cdot{\sin^2{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\left(\frac{1-\sin{\frac12\beta}}{2}\right).\tag{3}$$

Now, applying the sine law once more, this time in triangles $AFO$ and $ADF$, we obtain

$$\sin{\delta}=\frac{AO\cdot{\sin{\frac12(\alpha+\beta)}}}{AF},\tag{4}$$

$$\sin{\epsilon}=\frac{AD\cdot{\sin{\frac14(3\pi+\beta})}}{AF}.\tag{5}$$

Dividing $(4)$ by $(5)$ and then squaring, we have

$$\frac{\sin^2{\delta}}{\sin^2{\epsilon}}=\frac{\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}}{\sin^2{\frac14(3\pi+\beta})}=\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}\left(\frac{2}{1-\sin{\frac12\beta}}\right).\tag{6}$$

Substituting $(6)$ in $(3)$ and taking square roots,

$$\frac{GM}{DG}=\frac{\cos{\frac12\gamma}}{\cos{\frac12\alpha}}.$$

Analogously,

$$\frac{MN}{EN}=\frac{\cos{\frac12\alpha}}{\cos{\frac12\gamma}}.$$

This implies $\frac{GM}{DG}=\frac{EN}{MN}$ and, since $DM=ME$, we conclude that $DG=MN$.

$\square$