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 $M P=\frac{1}{2}, N T=2, B D=\sqrt{3}$. Find the area of triangle $A B C$.
Answer: (a) $90^{\circ};$ (b) $\frac{5 \sqrt{13}}{3 \sqrt{2}}$.

Solution. (a) Points $P$ and $T$ lie on the circle with diameter $BD$, so $\angle BPD = \angle BTD = 90^{\circ}$. Therefore, triangles $ADP$ and $DCT$ are right triangles; $PM$ and $TN$ are their medians. Since the median of a right triangle to the hypotenuse is half the hypotenuse, $TN = CN = DN$, $PM = AM = DM$. Let $\angle TCD = \gamma$. Since triangle $CTN$ is isosceles and $\angle CTN = \gamma$, $\angle TND = 2\gamma$ (as the external angle of $\triangle CTN$). Angles $\angle PMA$ and $\angle TND$ are equal due to the parallelism of lines $PM$ and $TN$. Since triangle $AMP$ is also isosceles, $\angle PAM = 90^{\circ} - \frac{1}{2} \angle PMA = 90^{\circ} - \frac{1}{2} \angle TND = 90^{\circ} - \gamma$. Therefore, the sum of angles $A$ and $C$ in triangle $ABC$ is $90^{\circ}$, and its third angle $\angle ABC$ is also $90^{\circ}$.

(b) As stated above, $CD = 2NT = 4$, $AD = 2MP = 1$. Let $\angle ADB = \psi$. Then $\angle BDC = 180^{\circ} - \psi$. By the cosine theorem for triangles $ABD$ and $ACD$, we get $AB^2 = 1 + 3 - 2\sqrt{3} \cos \psi$, $BC^2 = 16 + 3 - 8\sqrt{3} \cos (180^{\circ} - \psi)$. By the Pythagorean theorem, $AB^2 + BC^2 = AC^2 = 25$, from which it follows that $4 - 2\sqrt{3} \cos \psi + 19 + 8\sqrt{3} \cos \psi = 25$, $\cos \psi = \frac{1}{3\sqrt{3}}$. Next, we find: $\sin \psi = \sqrt{1 - \cos^2 \psi} = \frac{\sqrt{26}}{3\sqrt{3}}$, $S_{\triangle ABC} = S_{\triangle ABD} + S_{\triangle BCD} = \frac{1}{2} DA \cdot DB \sin \psi + \frac{1}{2} DC \cdot DB \sin (180^{\circ} - \psi) = \frac{1}{2} AC \cdot BD \sin \psi = \frac{1}{2} \cdot 5 \cdot \sqrt{3} \cdot \frac{\sqrt{26}}{3\sqrt{3}} = \frac{5\sqrt{13}}{3\sqrt{2}}$.