Answer:
see below
Step-by-step explanation:
Proposition:
Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE
Proof:
1)The lines AD and BC are parallel and AC their transversal therefore,
[tex] \displaystyle \angle DAC = \angle ACB \\ \ \qquad [\text{ alternate angles theorem}][/tex]
2)The lines AB and DC are parallel and BD their transversal therefore,
[tex] \displaystyle \angle BD C= \angle ABD \\ \ \qquad [\text{ alternate angles theorem}][/tex]
3)now in triangle ∆AEB and ∆CED
- [tex] \displaystyle \angle EAD=\angle ECB[/tex]
- [tex]\angle EDA=\angle EBC[/tex]
- [tex] \displaystyle AD=BC[/tex]
therefore,
[tex] \displaystyle \Delta AEB \cong \Delta CED [/tex]
hence,
Proven