Prove that the segment joining the midpoints of any two sides of a triangle, is parallel to the third side and is equal to half of the length of the third side.
Answers
Given :
A triangle ABC,E and F are mid points of side AB and AC respectively.
To prove :
line joining the mid points is parellel to the third side => EF || BC
and,
line joinining the mid points = half of third line
=> EF = 1/2 BC
Construction :
Through C, draw a line parellel to BA to meet EF produced at D.
statements reasons
In triangle AEF and triangle CDF
1. AF = CF 1. F is mid point of AC(given)
2. angle AFE = angle CFD 2. Vertically opposite angles
3.angle EAF = angle DCF 3.Alternate angles, BA || CD(by construction) and AC is a transversa.
4.triangle AEF congruent to 4. ASA rule of congruency
triangle CDF
5.EF = FD and AE = CD 5. c.p.c.t.
6.AE = BE 6. E is the mid point of AB (given)
7.BE = CD 7. From 5 and 6
8.EBCD is a parallelogram 8.BA || CD (construction)
and BE = CD (given)
9.EF || BC and ED = BC 9.Since EBCD is a parallelogram
10. EF = 1/2 ED 10.Since EF = FD,from 5
11.EF = 1/2 BC 11.Since ED = BC , from 9
Hence, EF || BC and EF = 1/2 BC.
Follow me