# prove that in a triangle the line segment joining the mid points of any to sides is parallel to the third side.  • 44

“The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half the third side.

This can be proved as: Also. EF = DF and EF + DF = ED = BC.
∴ 2EF = BC
⇒ EF = ½ BC

Hence, the mid-point theorem is proved
HOPE THIS HELPS!!!!
• 18 Given : E and F are the midpoints of AB and AC respectively.

CD II AB

in triangles AEF and CDF,

angle AFE = angle CFD      (vertically opposite angles)

AF = CF                                   (F is the midpoint of AC)

angle EAF = angle FCD        (alternate interior angles)

therefore, triangles AEF and CDF are congruent by ASA  congruence.

therefore, EF = DF (CPCT)

AE = EB = CD     (CPCT)

EB = CD  and CD II BA

therefore, BCDE is a parallelogram            (since a quadrilateral is a parallelogram if a pair of opposite sides are equal

and parallel)

therefore, EF II BC

hence, Proved.

hope it helped   :)

• 9
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