$ABCD$ 是一个菱形，且 $RD = CS$。 $SCB$、$DCR$ 和 $AQR$ 是直线。 证明三角形 $SCD$ 与 $RDA$ 全等。

1. Identify equal corresponding sides from the given condition and the rhombus property.

Given that

$RD = CS$.

Since $ABCD$ is a rhombus, all its sides are equal. Hence

$CD = DA$.

2. Show the included angles are equal using collinearity and parallel sides of the rhombus.

Since $S,C,B$ are collinear and $AD \parallel CB$, we have

$CS \parallel AD$.

Since $D,C,R$ are collinear, $CD$ and $RD$ lie on the same straight line.

Therefore,

$\angle SCD = \angle RDA$.

3. Conclude triangle congruence by SAS.

In $\triangle SCD$ and $\triangle RDA$,

$CS = RD$, $CD = DA$, and $\angle SCD = \angle RDA$.

Hence

$\triangle SCD \cong \triangle RDA \quad \text{(SAS)}$.