Given data:
The given triangle in which AD is on perpendicular bisector on BC.
In triangle ABD and ACD.
[tex]\begin{gathered} \angle ADB=\angle\text{ADC}=90^{\circ} \\ BD=CD(\text{given)} \\ AD=AD\text{ (common)} \\ \Delta ABD\cong\Delta ACD(\text{SAS)} \end{gathered}[/tex]
Simmilary triangle BED and triangle CED.
[tex]\begin{gathered} \angle BDE=\angle CDE \\ BD=CD \\ ED=ED \\ \Delta BED\cong\Delta CED(SAS) \end{gathered}[/tex]
The fisr expression can be written as,
[tex]\begin{gathered} \Delta ABD\cong\Delta ACD \\ \Delta\text{ABE}+\Delta BED\cong\Delta ACE+\Delta\text{CED} \end{gathered}[/tex]
Substitute CED in place of BED.
[tex]\begin{gathered} \Delta ABE+\Delta CED\cong\Delta ACE+\Delta CED \\ \Delta ABE\cong\Delta ACE \end{gathered}[/tex]
Thus, the triangle ABE is congruent to trriangle ACE.
