126,777 views
31 votes
31 votes
B>DGiven:. E is the midpoint of ADE is the midpoint of BCProve: ΔΑΕΒΑ ΔDECE is the midpoint of ADGiven

B>DGiven:. E is the midpoint of ADE is the midpoint of BCProve: ΔΑΕΒΑ ΔDECE is-example-1
B>DGiven:. E is the midpoint of ADE is the midpoint of BCProve: ΔΑΕΒΑ ΔDECE is-example-1
B>DGiven:. E is the midpoint of ADE is the midpoint of BCProve: ΔΑΕΒΑ ΔDECE is-example-2
User NeoNe
by
2.6k points

1 Answer

17 votes
17 votes

We are given a mid-point for segments AD and BC, we have the following:

segments AE and DE are congruent, that is:


AE\cong DE

By definition of mid-point.

Segments BE and CE are congruent, that is:


BE\cong CE

By definition of mid-point.

We also have that angles AEB and DEC are congruent, that is:


\angle AEB\cong\angle DEC

By the vertical angles theorem, which states that when two lines intercept their vertical or opposite angles are equal or congruent.

Now we can conclude that triangles AEB and DEC are congruent, that is:


\Delta AEB\cong\Delta DEC

Due to the Side-Angle-Side Theorem, which states that when two triangles have two congruent sides and the angle between the congruent sides also congruent, then the triangles are congruent.

User Ashley Grant
by
2.6k points