Question

Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark
Proof: 2 Marks


In trapezium, $ABCD,AB\parallel CD$ and $P,Q$ are the midpoints of $AC$ and $BD$

We need to prove $PQ\parallel AB$ and $DC$, $PQ=\frac{1}{2}(AB–DC)$

Construction: Join $DP$ and produce $DP$ to meet $AB$ in $R$.

Proof: Since $AB\parallel DC$ and transversal $AC$ cuts them.

$\angle 1=\angle 2$ …..(i) [Alternate interiror angles are equal]

Now, in $\Delta APR$ and $\Delta DPC$, we have

$\angle 1=\angle 2$ [From (i)]

$AP=CP$ [P is the mid-point of AC]

And, $\angle 3=\angle 4$ [Vertically opposite angles]

$\Rightarrow \Delta APR\cong \Delta CPD$ [ASA criterion of congruence]

$\Rightarrow AR=DC$ and $PR=DP$ [CPCT] …….(ii)

In $\Delta DRB,P$ and $Q$ are the mid points of sides $DR$ and $DB$ respectively

$\Rightarrow PQ\parallel RB$

$\Rightarrow PQ\parallel AB$ [RB is a part of AB]

$PQ\parallel AB$ and $PQ\parallel DC$ [AB || DC] (Given)]

Again,$P$ and $Q$ are the mid- point of sides $DR$ and $DB$ respectively in $\Delta DRB$.

$PQ=\frac{1}{2}RB$

$\Rightarrow PQ=\frac{1}{2}(AB–AR)$

$\Rightarrow PQ=\frac{1}{2}(AB–DC)$ [From (ii), $AR=DC$]