Question

Given: 1 = 2 3 = 4 D midpoint of BC = DE Prove: A = E 1. ∠1=∠2, ∠3=∠4, D is midpoint of segment BE, BC = DE CPCTE 2. BD = DE Given 3. BC = BD Substitution 4. Triangle ABD congruent to Triangle EBC Definition of midpoint 5. ∠A = ∠E ASA

Answer 1

Answers:

1. <1=<2, <3=<4, D is midpoint of segment BE, BC=DE → Given

2. BD=DE → Definition of midpoint

3. BC=BD → Substitution property

4. Triangle ABD congruent to triangle EBC → ASA

5. <A=<E → CPCTE

Solution:

1. <1=<2, <3=<4, D is midpoint of segment BE, BC=DE

All this information is given by the problem.

2. BD=DE

If D is the midpoint of segment BE, D divides this segment into two congruent parts BD and DE, then BD must be equal to DE: BD=DE by definition of midpoint.

3. BC=BD

The problem says that BC=DE (1)

And by point 2 we know that BD=DE→DE=BD (2)

Then by substitution property if we can replace DE in equation (1) by BD (because of equation (2) ):

(1) BC=DE and (2) DE=BD → (1) BC=BD

4. Triangle ABD congruent to triangle EBC

The triangles ABD and EBC have a congruent side (BD in triangle ABD and BC in triangle EBC) and the two adjacent angles congruent too (<1 in triangle ABD with <2 in triangle EBC; and <3 in triangle ABD with <4 in triangle EBC), then by Angle Side Angle (ASA) the two triangles must be congruent.

5. <A=<E

If the triangles are congruent (by point 4), all its parts must be congruent too (Corresponding Parts of Congruent Triangles are Equal: CPCTE), then the third angle in triangle ABD (<A) must be equal to the third angle in triangle EBC (<E):

<A=<E