Question

prove that the line segment joining the mid points of any two sides of a triangle is parallel to the third side​

Answer 1

Answer:

Consider the triangle ABC, as shown in the figure below. Let E and D be the midpoints of the sides AC and AB respectively. Then the line DE is said to be parallel to the side BC, whereas the side DE is half of the side BC, i.e.

DE || BC

So, DE = 1/2 × BC

Given: D and E are the mid-points of sides AB and AC of ΔABC respectively.

Construction: In ΔABC, through C, draw a line parallel to BA, and extend DE such that it meets this parallel line at F.

Proof: Compare ΔAED with ΔCEF:

AE = EC (E is the midpoint of AC)

∠DAE = ∠FCE (alternate interior angles)

∠DEA = ∠FEC (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,

DF || BC ⇒ DE || BC

and, DF = BC

⇒ DE + EF = BC

⇒ 2DE = BC (as, DE = EF, proved above)

⇒ DE = 1/2 × BC

Thus proved