4. In triangle $K L M$, medians $L D$ and $M E$ intersect at point $G$. The circle constructed on segment $L G$ as a diameter passes through vertex $M$ and is tangent to line $D E$. It is known that $G M=6$. Find the height $K T$ of triangle $K L M$, the angle $L G M$, and the area of triangle $K L M$.
Answer: $K T=18, \angle L G M=60^{\circ}, S_{K L M}=54 \sqrt{3}$.

Solution. Let the center of the circle be denoted as $O$, the point of tangency of the 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, $D G=\frac{1}{2} L G=R, M E=\frac{3}{2} G M=9$. Note that in the right triangle $O D S$, the leg $O S=R$, and the hypotenuse $O D=2 R$. Therefore, $\angle S D O=30^{\circ}$. The segment $D E$ is the midline of triangle $K L M$, so $D E \| L M$, $\angle G L M=\angle G D E=30^{\circ}$. From triangle $G L M$, we find that $L M=M G \operatorname{ctg} 30^{\circ}=6 \sqrt{3}, \angle L G M=60^{\circ}$.

The height $K T$ of the original triangle is twice the length of $M E$ (since $M E$ is the midline of triangle $K L T$). Thus, $K T=18$. The area of the triangle is $\frac{1}{2} K T \cdot L M=54 \sqrt{3}$.