Question

Question 16
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.

Answer 1

Since M is the mid-point of AB,
AM = MB
Now, in $\Delta AMNand\Delta BMN,AM=MB$ [proved above]
$\angle 3=\angle 4$ [each ${90}^{\circ }$]
MN = MN [common side]
$\therefore \Delta AMN\cong BMN$ [by SAS congruence rule]
$\therefore \angle 1=\angle 2$ [by CPCT]
On multiplying both sides of above equation by - 1 and than adding ${90}^{\circ }$ both sides, we get ${90}^{\circ }-\angle 1={90}^{\circ }-\angle 2$
$\Rightarrow \angle AND=\angle BNC...\left(i\right)$

Now, in $\Delta ADNand\Delta BCN,$
$\angle AND=\angle BNC$ [from Eq.(i)]
AN = BN $[\because \Delta AMN\cong \Delta BMN]$
and DN = NC $[\therefore$ N is the mid-point of CD (given)]
$\therefore \Delta ADN\cong \Delta BCN$ [by SAS congruence rule]
Hence, AD = BC [by CPCT]