Question

The answer has to be a geometric proof. Thank you!

Answer 1

Given data:

The given triangle in which AD is on perpendicular bisector on BC.

In triangle ABD and ACD.

$\begin{gathered} \angle ADB=\angle\text{ADC}=90^{\circ} \\ BD=CD(\text{given)} \\ AD=AD\text{ (common)} \\ \Delta ABD\cong\Delta ACD(\text{SAS)} \end{gathered}$

Simmilary triangle BED and triangle CED.

$\begin{gathered} \angle BDE=\angle CDE \\ BD=CD \\ ED=ED \\ \Delta BED\cong\Delta CED(SAS) \end{gathered}$

The fisr expression can be written as,

$\begin{gathered} \Delta ABD\cong\Delta ACD \\ \Delta\text{ABE}+\Delta BED\cong\Delta ACE+\Delta\text{CED} \end{gathered}$

Substitute CED in place of BED.

$\begin{gathered} \Delta ABE+\Delta CED\cong\Delta ACE+\Delta CED \\ \Delta ABE\cong\Delta ACE \end{gathered}$

Thus, the triangle ABE is congruent to trriangle ACE.