ABCD 是正方形，且 AQ 垂直于 PR。 PQR 和 ADR 是直线。 DQ : DC = 1 : 2。 证明三角形 DQR 与三角形 CQP 全等。
为你作出的每个陈述说明理由。 答案

1. Show that the corresponding side lengths $DQ$ and $QC$ are equal.

Given $DQ:DC=1:2$,
$DQ=\frac12DC.$
Since $Q$ lies on $DC$,
$DQ+QC=DC.$
Hence
$QC=DC-DQ=DC-\frac12DC=\frac12DC.$
Therefore,
$DQ=QC.$

2. Show that two pairs of corresponding angles are equal and conclude congruence.

In the square, $AD\perp DC$ and $BC\perp DC$. Since $D,A,R$ are collinear and $C,B,P$ are collinear,
$\angle QDR=\angle QCP=90^\circ.$
Also, since $D,Q,C$ are collinear and $R,Q,P$ are collinear,
$\angle DQR=\angle CQP$
vertically opposite angles.

Thus, with $DQ=QC$,
$\triangle DQR\cong\triangle CQP$
by ASA.