# Congruent Triangles Geometry Maths Tuition: Solution

Solution:

(a) $\angle EBD=\angle BEC$ (given)

BE is a common side for both triangles $\triangle BCE$ and $\triangle EDB$

BD=EC (given)

Therefore, $\triangle BCE \equiv \triangle EDB$ (SAS)

(proved)

(b)

Since $\triangle BCE\equiv\triangle EDB$ we have $\angle CBE=\triangle DEB$

Thus, $\begin{array}{rcl}\angle ABE&=&180^\circ - \angle CBE\\ &=&180^\circ -\angle DEB\\ &=& \angle AEB \end{array}$

Therefore, $\triangle ABE$ is an isosceles triangle.

Thus, $AB=AE$

(proved)

## Author: mathtuition88

https://mathtuition88.com/

This site uses Akismet to reduce spam. Learn how your comment data is processed.