# Midsegment theorem of trapezium

In this chapter we will understand and prove the midsegment theorem of trapezium.

In the proof, we have extensively used the concept of midpoint theorem of triangle. So make sure to understand that theorem before moving on with this chapter.

Note that this theorem is also called midpoint theorem of trapezium.

## What is midpoint theorem of trapezium ?

The theorem states that in trapezium, the length of line formed by joining midpoint of two sides and parallel to third side is equal to half of sum of parallel sides.

Given above is the trapezium ABCD in which;

⟹ side AB is parallel to CD.
⟹ E is the mid point of side AD.
⟹ F is the midpoint of side BC.

EF is the line segment formed by joining points E & F and is parallel to sides AB & CD.

According to the midsegment theorem of trapezium, the length of side EF is equal to half of the sum pf parallel sides AB & CD.

\mathtt{EF\ =\ \frac{1}{2}( AB\ +\ CD)}

### Proof of midsegment theorem of trapezium

Given:
Given above is trapezium ABCD in which AB || CD.

Here E and F are the midpoints of sides AD and BC respectively.

Line EF is drawn parallel to parallel to side AB and CD.

Construction:
Extend line CD to meet line AF at point O.

To Prove:
\mathtt{EF\ =\ \frac{1}{2}( AB\ +\ CD)}

Proof
Take triangle ABF and COF.

∠AFB = ∠CFO { vertically opposite angles }
∠ABF = ∠OCF { alternate angles }
BF = FC { F is midpoint of BC }

By ASA congruency condition, both triangles are congruent.
\mathtt{\triangle ABF\ \cong \triangle COF}

Since both triangles are congruent we can say that;
(i) AF = FO
It means that F is midpoint of AO

(ii) AB = CO.

We know that;
F is midpoint of AO.

Using midpoint theorem of triangle, we can say that;

\mathtt{EF\ =\ \frac{1}{2}( DO)}\\\ \\ \mathtt{EF\ =\ \frac{1}{2}( DC+CO)}\\\ \\ \mathtt{As,\ AB\ =\ CO;}\\\ \\ \mathtt{EF\ =\ \frac{1}{2}( DC+AB)}

Hence, we proved that length of EF is half of sum of side AB and CD.

### Questions on midsegment theorem of trapezium

Example 01
Given below is trapezium PQRS in which M and N is the midpoint of side PS and RQ respectively. Also side PQ || MN || SR. If PQ = 10 cm and MN = 8 cm. Find the length of side SR.

Solution
Using the midsegment theorem of trapezium.

\mathtt{MN\ =\ \frac{1}{2}( PQ+SR)}

Putting the values, we get;

\mathtt{8\ =\ \frac{1}{2}( 10\ +\ SR)}\\\ \\ \mathtt{16\ =\ 10\ +\ SR}\\\ \\ \mathtt{SR\ =\ 6\ cm\ }

Hence, length of SR is 6 cm.

Example 02
Given below is trapezium ABCD . The line RS is formed by joining midpoints of respective sides. If sides AB || RS || CD, then find length of segment RS.

Solution
Using the midpoint theorem of trapezium.

\mathtt{RS\ =\ \frac{1}{2}( AB+CD)}

Putting the values;

\mathtt{RS\ =\ \frac{1}{2}( AB+CD)}\\\ \\ \mathtt{RS\ =\ \frac{1}{2}( 8\ +\ 5)}\\\ \\ \mathtt{RS=\ \frac{13}{2}}\\\ \\ \mathtt{RS\ =\ 6.5\ cm}

Hence, length of RS is 6.5 cm