Solve Converse of Mid-point Theorem Problems

In this article we are going to solve converse of mid-point theorem problems. The straight line which is drawn through the mid point of one side of a triangle is parallel to another side bisects the third side. This is the converse of mid-point theorem. But the mid point theorem states that the line segments joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.

Solve Converse of Mid-point Theorem Problems:

Given: In Triangle ABC, D is the midpoint of AB and DE is drawn parallel to BC.

To prove: AE = EC

Construction: Draw CF parallel to BA to meet DB proceed at F.




Proof : 

Statement	Reason
DB || BC BD || CF BCFD is a parallelogram Therefore BD = CF ... ( i ) BD = AD ... ( ii ) Therefore AD = CF In triangle ADE and CFE, we have AD = CF Angle ADE = Angle CFE Angle AED = Angle CEF Therefore triangle ADE = Triangle CFE Therefore AE = EC Therefore DE bisects AC Hence proved	Given By construction Both pairs of opposite sides are parallel Opposite sides of a parallelogram unequal D is the mid point of AB From ( i ) and ( ii ) Just proved Alternate to angle S Vertically opposite angles of S Angle Angle Side criterion C. P. T. C.

Example Problem - Solve Converse of Mid-point Theorem Problems:

ABC of a right angled triangle at B, Here the mid-point of AC is P.

Solve that PB = PA = ½


Solution for the  converse of mid-point theorem problems:

Given:

ABC of a right angled triangle at B, Here the mid-point of AC is P.

To prove:

PB = PA = `1/2` AC.

Construction:

Through P, draw a line parallel to BC, meeting AB at Q.

Proof:

AQP = ABC (corresponding angles)

AQP = 90 degree.

ABC = 90 degree

In D APQ and D BPQ

AQ = BQ

AQP = BQP = 90 degree

PQ = PQ

Therefore in triangle APQ = Triangle BPQ

PA = PB (corresponding parts of congruent triangles)

PA = PB = `1/2` AC

PA = `1/2` AC (given)