98. The height dropped from the vertex of the right angle of a right-angled triangle to the hypotenuse divides the triangle into two triangles, each of which has an inscribed circle. Determine the angles and the area of the triangle formed by the legs of the original triangle and the line passing through the centers of the circles, if the height of the original triangle is $h$.
98. Let $ABC$ be a given triangle, $CD$ - the altitude, $O_{1}$ and $O_{2}$ the centers of the circles inscribed in $\triangle ACD$ and $\triangle BDC$, $K$ and $L$ the points of intersection of the lines $DO_{1}$ and $DO_{2}$ with $AC$ and $CB$. Since $\triangle ADC$ is similar to $\triangle CDB$, and $KD$ and $LD$ are the bisectors of the right angles of these triangles, $O_{1}$ and $O_{2}$ divide $KD$ and $LD$ in the same ratio. Therefore, $KL \parallel O_{1}O_{2}$. But the quadrilateral $CKDL$ is inscribed $\left(\widehat{KCL}=\widehat{KDL}=90^{\circ}\right)$. Hence, $\widehat{CKL}=\widehat{CDL}=\frac{\pi}{4}, \widehat{CLK}=\widehat{CDK}=\frac{\pi}{4}$. Thus, the line $O_{1}O_{2}$ forms angles of $\frac{\pi}{4}$ with the legs. If $M$ and $N$ are the points of intersection of $O_{1}O_{2}$ with $CB$ and $AC$, then $\triangle CMO_{2}=\triangle CDO_{2}\left(CO_{2}\right.$ - common, $\widehat{O_{2}CD}=\widehat{O_{2}CM}, \widehat{CDO_{2}}=\frac{\pi}{4}=\widehat{CMO_{2}})$. Therefore, $|CM|=|NC|=h$.

Answer: The area of the triangle is $\frac{h^{2}}{2}$, the angles are $\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{2}$.