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 $M P=\frac{1}{2}, N T=\frac{5}{2}, B D=2$. Find the area of triangle $A B C$.
Answer: (a) $90^{\circ} ;$ (b) $\frac{9 \sqrt{7}}{4}$.

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 half 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, $C D = 2 N T = 5$, $A D = 2 M P = 1$. 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 $A B^{2} = 4 + 1 - 4 \cos \psi$, $B C^{2} = 25 + 4 - 20 \cos (180^{\circ} - \psi)$. By the Pythagorean theorem, $A B^{2} + B C^{2} = A C^{2} = 36$, from which it follows that $5 - 4 \cos \psi + 29 + 20 \cos \psi = 36$, $\cos \psi = \frac{1}{8}$. Further, we find: $\sin \psi = \sqrt{1 - \cos^{2} \psi} = \frac{3 \sqrt{7}}{8}$, $S_{\triangle A B C} = S_{\triangle A B D} + S_{\triangle B C D} = \frac{1}{2} D A \cdot D B \sin \psi + \frac{1}{2} D C \cdot D B \sin (180^{\circ} - \psi) = \frac{1}{2} A C \cdot B D \sin \psi = \frac{1}{2} \cdot 6 \cdot 2 \cdot \frac{3 \sqrt{7}}{8} = \frac{9 \sqrt{7}}{4}$.