1. [5 points] Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$.

a) Find the angle $A B C$.

b) Suppose additionally that $M P=\frac{1}{2}, N T=1, B D=\frac{4}{3}$. Find the area of triangle $A B C$.
Answer: (a) $90^{\circ} ;$ (b) $\frac{\sqrt{35}}{3}$.

Solution. (a) Points $P$ and $T$ lie on the circle with diameter $B D$, so $\angle B P D=\angle B T D=90^{\circ}$. Therefore, triangles $A D P$ and $D C T$ are right triangles; $P M$ and $T N$ are their medians. Since the median of a right triangle to the hypotenuse is equal to half of the hypotenuse, $T N=C N=D N, P M=$ $A M=D M$. Let $\angle T C D=\gamma$. Since triangle $C T N$ is isosceles, and $\angle C T N=\gamma, \angle T N D=$ $2 \gamma$ (as the exterior angle of $\triangle C T N$). Angles $P M A$ and $T N D$ are equal due to the parallelism of lines $P M$ and $T N$. Since triangle $A M P$ is also isosceles, $\angle P A M=90^{\circ}-\frac{1}{2} \angle P M A=90^{\circ}-\frac{1}{2} \angle T N D=$ $90^{\circ}-\gamma$. Therefore, the sum of angles $A$ and $C$ of triangle $A B C$ is $90^{\circ}$, and its third angle $\angle A B C$ is also $90^{\circ}$.

(b) As stated above, $C D=2 N T=2, A D=2 M P=1$. Let $\angle A D B=\psi$. Then $\angle B D C=$ $180^{\circ}-\psi$. By the cosine theorem for triangles $A B D$ and $A C D$, we get $A B^{2}=\frac{16}{9}+1-\frac{8}{3} \cos \psi$, $B C^{2}=\frac{16}{9}+4-\frac{16}{3} \cos \left(180^{\circ}-\psi\right)$. But by the Pythagorean theorem, $A B^{2}+B C^{2}=A C^{2}=9$, from which it follows that $\frac{25}{9}-\frac{8}{3} \cos \psi+\frac{52}{9}+\frac{16}{3} \cos \psi=9, \cos \psi=\frac{1}{6}$. Further, we find: $\sin \psi=\sqrt{1-\cos ^{2} \psi}=\frac{\sqrt{35}}{6}, S_{\triangle A B C}=$ $S_{\triangle A B D}+S_{\triangle B C D}=\frac{1}{2} D A \cdot D B \sin \psi+\frac{1}{2} D C \cdot D B \sin \left(180^{\circ}-\psi\right)=\frac{1}{2} A C \cdot B D \sin \psi=\frac{1}{2} \cdot 3 \cdot \frac{4}{3} \cdot \frac{\sqrt{35}}{6}=\frac{\sqrt{35}}{3}$.