Question

Show by construction that the segment that connects the midpoint of two sides of triangle `DeltaABC` is parallel to the third and half it's length?

Answer 1

See the Proof in Explanation.

Explanation:

The construction : Complete the parallelogram `ABFC`, &, extend

seg. `MN` to meet `BF` in `G`.

In `DeltaCMN` and `DeltaBGN`,

`m/_MCN=m/_GBN...".[line "AMC ||" line "BG]`

`m/_MNC=m/_GNB........[Opp. /_s]`

`CN=BN......[N mid-pt. of BC]`

`:. DeltaCMN~=DeltaBGN`.

`:. CM=BG." But, as, "M" is mid-pt. of "AC, CM=MA`.

`:. BG=AM. As, BG || AM," this means that, "ABGM" is a "||grm`.

Hence, `MN || AB, and, MG=AB`

Since, `DeltaCMN~=DeltaBGN, MN=NG, or, MN=1/2MG=1/2AB`

Hence, the Proof.

Enjoy Maths.!