We can assume that, $AB=4$ and $BC=5$. By Cosine Law, $AC=\sqrt{21}$, and by Pythagorean Theorem, $BD=\frac{10}{\sqrt7}$, which implies that $MD=\frac{5}{\sqrt7}$. Also, $AD=\frac{2\sqrt3}{\sqrt7},CD=\frac{5\sqrt3}{\sqrt7}$. From $\frac{MD}{CD}=\frac{1}{\sqrt{3}}$, we have, $\angle DMC=60^\circ$. Because $BMNC$ is cyclic, we have $\angle MBN=\angle MCN\implies\angle DBN=\angle DCM\implies\Delta DBN\sim\Delta DCM$. From this, we have $\frac{DB}{DN}=\frac{DC}{DM}\implies DN=\frac{10}{\sqrt{21}}$ and $\angle DNB=60^\circ$. Claim 1: $ED||BN$ Proof: It suffices to show that $\Delta AED\sim\Delta ABN$. It is well-known that $\Delta AED\sim\Delta ACB$, and because $\angle ABC=\angle DNB=60^\circ$, we now have $\Delta ACB\sim\Delta ABN$. Claim 2: $DF\perp BN$ Proof: We know that $BM:MD=1:1$ and $CN:ND=1:2$. Using mass points tells us that $MF:FC=1:1$. This means $F$ is the midpoint of $MC$, and because $MDC$ is a right triangle, $DF=MF=\frac{1}{2}MC=\frac{5}{\sqrt7}$ We know that $\angle DBF=30^\circ$, and by using Sine Law, $\angle DFB=90^\circ$. From the two claims, we have $\angle EDP=90^\circ$.

Let $AB=4$ and $BC=5$, since we can resize triangle to right size. By the law of cosines, $AC=AB^2+BC^2-2BC\cdot AB\cdot \cos{\angle ABC}=16+25-4\cdot 5=21\implies AC=\sqrt{21}$. By area calculation, $AC\cdot BD=AB\cdot CB\cdot \sin{\angle ABC}\implies BD=\frac{\sqrt{3}\cdot 2\cdot 5}{4\sqrt{21}}=\frac{10\sqrt{7}}{7}$. $DM\cdot DB=DN\cdot DC\implies DN=\frac{DM\cdot DB}{DC}=\frac{DM\cdot DB}{DC}$ $$\cos{\angle BCA}=\frac{CB^2+AC^2-AB^2}{2 \cdot CA\cdot CB}=\frac{25+21-16}{2 \cdot \sqrt{21}\cdot 5}=\frac{\sqrt 3}{\sqrt 7}$$Thus, $$\cos{\angle BCA}=\frac{CD}{BC}\implies CD=\cos{\angle BCA}\cdot BC=\frac{\sqrt 3}{\sqrt 7}\cdot 5=\frac{\sqrt{3}\cdot 5}{\sqrt 7}$$Thus, $$DN=\frac{\frac{5\sqrt{7}}{7}\cdot \frac{10\sqrt{7}}{7}}{\frac{\sqrt{3}\cdot 5}{\sqrt 7}}=\frac{10\cdot \sqrt{7}}{7\sqrt{3}}$$Thus, $$\tan{\angle DBN}=\frac{DN}{BD}=\frac{\frac{10\cdot \sqrt{7}}{7\sqrt{3}}}{\frac{10\sqrt{7}}{7}}=\frac{1}{\sqrt{3}}\implies \angle DBN=30^{\circ}.$$Hence, $\angle DBN=\angle MBN=\angle PCN=30^{\circ}$ and $\angle BND=60^{\circ}$, thus $NC=PN$. We have $$PN=NC=DC-DN=\frac{\sqrt{3}\cdot 5}{\sqrt 7}-\frac{10}{\sqrt{7}\sqrt{3}}=\frac{1}{\sqrt{7}}\cdot \frac{5}{\sqrt{3}}$$Hence, we easily have that $\triangle DPN$ is right triangle. Hence, $\triangle DPM$ is equilateral. Also, $$\cos{\angle BAC}=\frac{AB^2+AC^2-BC^2}{2 \cdot AB\cdot AC}=\frac{16+21-25}{2 \cdot \sqrt{21}\cdot 4}=\frac{3}{2\sqrt 21}$$Hence, $$AE=\cos{\angle BAC}\cdot AC=\frac{3}{2}$$and $$BE=\cos{\angle ABC}\cdot BC=\frac{1}{2}\cdot 5=\frac{5}{2}$$and $$AD=AC-CD=\sqrt{21}-\frac{\sqrt{3}\cdot 5}{\sqrt{7}}=\frac{7{\sqrt{3}}-\sqrt{3}\cdot 5}{\sqrt{7}}=\frac{2{\sqrt{3}}}{\sqrt{7}}.$$Thus, $$\frac{3}{5}=\frac{\frac{2{\sqrt{3}}}{\sqrt{7}}}{\frac{10\cdot \sqrt{7}}{7\sqrt{3}}}\Longleftrightarrow \frac{\frac{3}{2}}{\frac{5}{2}}=\frac{\frac{2{\sqrt{3}}}{\sqrt{7}}}{\frac{10\cdot \sqrt{7}}{7\sqrt{3}}}\Longleftrightarrow \frac{AE}{EB}=\frac{AD}{DN}\Longleftrightarrow \text{ $DE$ and $BN$ are parallel.}$$Hence, $$\angle EDP=\angle DPN=90^{\circ}.\qquad \square$$