Question

Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side.

Answer 1

Answer:

Through C, draw a line parallel to BA, and extend DE such that it meets this parallel at F, as shown below:

Midpoint Theorem

Compare

Δ

A

E

D

with

Δ

C

E

F

:

1. AE = EC (E is the midpoint of AC)

2.

∠

D

A

E

=

∠

F

C

E

(alternate interior angles)

3.

∠

D

E

A

=

∠

F

E

C

(vertically opposite angles)

By the ASA criterion, the two triangles are congruent. Thus, DE = EF and AD = CF. But AD is also equal to BD, which means that BD = CF (also, BD || CF by our construction). This implies that BCFD is a parallelogram. Thus,

1. DF || BC è DE || BC

2. DE = EF = ½(DF) = ½(BC) èDE = ½(BC)

This completes our proof.

Answer 2

Answer:

Step-by-step explanation: