Problem 772

## Solution.

Let $E$ and $F$ be the points of intersection of segment $AN$ with $DM$ and $BP$, respectively, and $K$ and $L$ be the points of intersection of segment $CQ$ with $BP$ and $DM$, respectively (Fig. 10.74). $\triangle AQC = \triangle ANC$ (by two sides and the included angle), therefore, $\angle QCA = \angle NAC$, which means $AN \parallel CQ$. Similarly, $DM \parallel BP$. Thus, $EFKL$ is a parallelogram. $S_{\triangle ACD} = \frac{1}{2} S_{ABCD} = 50$. $S_{\triangle CQD} = \frac{1}{2} S_{\triangle ACD} = 25$. Since $PD = PC$ and $BP \parallel DM$, then $CK = LK$, similarly $DL = EL$, $AE = FE$. Therefore, $QL$ is the midline of $\triangle ADE$, i.e., $QL = \frac{1}{2} AE = \frac{1}{2} EF = \frac{1}{2} KL$. Then $QL = \frac{1}{5} CQ$, $S_{\triangle DQL} = \frac{1}{5} S_{\triangle DCQ} = 5$. Let $\angle ELK = \alpha$, then $\angle DLQ = \alpha \cdot \frac{S_{EFLK}}{S_{DQL}} = \frac{EL \cdot LK \cdot \sin \alpha}{\frac{1}{2} DL \cdot QL \cdot \sin \alpha} = \frac{DL \cdot LK}{\frac{1}{4} DL \cdot LK} = 4$. Therefore, $S_{EFLK} = 4 S_{\triangle DQL} = 20$.

Answer: $20 \text{ cm}^2$.

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Fig. 10.75

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Fig. 10.76