在下图角中，ABCD 是一个菱形，对角线 AC 与 y 轴相交于点 E。 证明三角形 AEB 与三角形 AED 全等。 答 ___

1. State the equal sides from the definition of a rhombus.

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

2. State the equal segments from the diagonals of a rhombus bisecting each other.

The diagonals of a rhombus bisect each other. Since $AC$ and $BD$ intersect at $E$,
$BE=DE.$

3. Use the common side and conclude congruence by SSS.

Also,
$AE=AE$
(common side). Therefore, the three corresponding sides are equal:
$AB=AD,\quad BE=DE,\quad AE=AE.$
Hence
$\triangle AEB \cong \triangle AED \quad \text{by SSS}.$