In the given Figure, $□$ ABCD is a trapezium. AB || DC .Points P and Q are midpoints of seg AD and seg BC respectively.
Then prove that, PQ || AB and PQ = $\frac{1}{2}$ ( AB + DC ).

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Solution

Construction: Join PB and extend it to meet CD produced at R.
To prove: PQ || AB and PQ = $\frac{1}{2}$ ( AB + DC)
Proof: In $△$ ABP and $△$ DRP,
$∠ APB = ∠ DPR (Vertically opposite angles) ∠ PDA = ∠ PAB (Alternate interior angles are equal) AP = PD (P is the mid point of AD)$
Thus, by ASA congruency, $△$ ABP $≅$ $△$ DRP.
By CPCT, PB = PR and AB = RD.
In $△$ BRC,
Q is the mid point of BC (Given)
P is the mid point of BR (As PB = PR)
So, by midpoint theorem, PQ || RC
$\Rightarrow$ PQ || DC
But AB || DC (Given)
So, PQ || AB.
Also, PQ = $\frac{1}{2} (RC)$
$PQ = \frac{1}{2} (RD + DC) PQ = \frac{1}{2} (AB + DC) (∵ AB = RD)$

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