Questions: Given: AE is the perpendicular bisector of BL Prove: triangle AEB is congruent to triangle AEL.

Transcript text: Given: $\overline{A E}$ is the perpendicular bisector of $\overline{B L}$ Prove: $\triangle A E B \cap \triangle A E L$.

Solution

Solution Steps

Step 1: Analyze the given information

We are given that $\overline{AE}$ is the perpendicular bisector of $\overline{BL}$. This means that $AE \perp BL$ and $BE = EL$. Also, AE is a common side to both triangles $\triangle AEB$ and $\triangle AEL$.

Step 2: Identify congruent parts

From the given information, we know that:

$BE = EL$ (Since AE bisects BL)
$\angle AEB = \angle AEL = 90^\circ$ (Since AE is perpendicular to BL)
$AE = AE$ (Common side)

Step 3: Determine triangle congruence

Using the Side-Angle-Side (SAS) postulate, we can conclude that $\triangle AEB \cong \triangle AEL$. Since congruent triangles have identical vertices, sides, and angles, they share all their points.

Final Answer

$\boxed{\triangle AEB \cong \triangle AEL}$

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