4. In triangle $K L M$, medians $L D$ and $M E$ intersect at point $T$. The circle constructed on segment $M T$ as a diameter passes through vertex $L$ and is tangent to line $D E$. It is known that $D T=2$. Find the height $K H$ of triangle $K L M$, the angle $E T D$, and the area of triangle $K L M$.
Answer: $K H=12, \angle E T D=60^{\circ}, S_{K L M}=24 \sqrt{3}$.

Solution. Let the center of the circle be denoted as $O$, the point of tangency of line $D E$ with the circle as $S$, and the radius of the circle as $R$. Since the medians intersect at a point dividing them in the ratio $2:1$ from the vertex, $E T=\frac{1}{2} M T=R, L D=3 D T=6, L T=2 D T=4$. Note that in the right triangle $O E S$, the leg $O S=R$, and the hypotenuse $O E=2 R$. Therefore, $\angle S E O=30^{\circ}$. Segment $D E$ is the midline of triangle $K L M$, so $D E \| L M, \quad \angle T M L=\angle T E D=30^{\circ}$. From triangle $T L M$, we find that $L M=T L \operatorname{ctg} 30^{\circ}=4 \sqrt{3}$. $\angle E T D=\angle L T M=60^{\circ}$.

The height $K H$ of the original triangle is twice the length of $L D$ (since $L D$ is the midline of triangle $K M H$). Thus, $K H=12$. The area of the triangle is $\frac{1}{2} K H \cdot L M=24 \sqrt{3}$.