Question

In the adjacent figure, $\square$ABCD is a trapezium AB || DC . Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB .

Answer 1

Construction: Join MN. Also, join DM and produce it to AB in P.
To prove: MN || AB
Proof: AB || DC and AC is the transversal.
So, $\angle 1=\angle 2\left(\mathrm{Alternate}\mathrm{angles}\mathrm{are}\mathrm{equal}\right)$
In $△$AMP and $△$CMD,
$\angle 1=\angle 2\left(\mathrm{Alternate}\mathrm{angles}\mathrm{are}\mathrm{equal}\right)$
$\angle 3=\angle 4$(Vertically opposite angles)
AM = MC (M is midpoint of AC)
So, $△$AMP $\cong$ $△$CMD (ASA congruency criteria)
By CPCT, DM = MP.
So, M is the midpoint of DP.
In $△$DPB,
M is the midpoint of DP and N is the midpoint of DB.
By, midpoint theorem MN || AB.