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?

Question

Ella Armstrong · Accepted Answer

See the Proof in 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.!

Allison Allen · Answer

undefined To show that the segment connecting the midpoint of two sides of triangle $ \Delta ABC $ is parallel to the third side and half its length, follow these steps:

1. Draw triangle $ \Delta ABC $.
2. Mark the midpoints of two sides of the triangle. Let's call them $ M $ and $ N $, where $ M $ is the midpoint of side $ AB $ and $ N $ is the midpoint of side $ AC $.
3. Connect $ M $ and $ N $ to form segment $ MN $.
4. Draw a line parallel to side $ BC $ passing through point $ M $. Let this line intersect side $ AC $ at point $ P $.
5. Since $ MN $ is parallel to side $ BC $, angle $ AMP $ and angle $ ABC $ are corresponding angles and are therefore equal.
6. Also, since $ M $ is the midpoint of side $ AB $, $ AM = \frac{1}{2} AB $.
7. Similarly, since $ N $ is the midpoint of side $ AC $, $ AN = \frac{1}{2} AC $.
8. Using the midpoint theorem, segment $ MN $ is parallel to side $ BC $ and $ MN = \frac{1}{2} BC $.

Therefore, segment $ MN $ is parallel to side $ BC $ and half its length.