10.198. The bisectors of the obtuse angles at the base of the trapezoid intersect on the other base of the trapezoid. Find all sides of the trapezoid if its height is 12 cm, and the lengths of the bisectors are 15 and 13 cm.
## Solution.

Let in trapezoid $A B C K$ (Fig. 10.8) $B C \| A K, B D$ and $C D$ be the bisectors of $\angle A B C$ and $\angle B C K, D E$ - the height, $D E=12$ cm, $B D=15$ cm, $C D=13$ cm. From $\triangle B E D \quad\left(\angle B E D=90^{\circ}\right): B E=\sqrt{B D^{2}-D E^{2}}=9$ cm. From the right

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0715.jpg?height=458&width=642&top_left_y=78&top_left_x=93)

Fig. 10.8

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0715.jpg?height=468&width=577&top_left_y=65&top_left_x=686)

Fig. 10.9

triangle $\triangle C E D \quad C E=\sqrt{C D^{2}-D E^{2}}=5 \text{ cm}$. Then $B C=B E+C E=14$ cm. $\angle 1=\angle 3$ - by condition, $\angle 1=\angle 2$ - alternate interior angles when $B C \| A D$ and the transversal $B D$. Therefore, $\angle 2=\angle 3$, $A D=A B$. Similarly, $\angle 5=\angle 6, C K=D K$. $B M$ and $C N$ are the heights of the trapezoid. $B M=C N=D E=12$ cm, hence $M D=B E=9$ cm, $D N=E C=5$ cm. Let $A B=x$ cm, then $A M=(x-9)$ cm. From the right triangle $\triangle A M B \quad A B^{2}=A M^{2}+B M^{2} ; x^{2}=$ $=(x-9)^{2}+144, x=12.5$. Therefore, $A D=A B=12.5$ cm. Let $C K=y$ cm. Then $N K=(y-5)$ cm. From $\triangle C N K \quad\left(\angle C N K=90^{\circ}\right)$ : $C K^{2}=C N^{2}+N K^{2} ; y^{2}=(y-5)^{2}+144 ; y=16.9$. Therefore, $D K=C K=16.9$ cm. Then $A K=A D+D K=29.4$ cm.

Answer: 29.4 cm; 12.5 cm; 14 cm; 16.9 cm.