# The line segments joining the midpoints M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC.

Question:

The line segments joining the midpoints and of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.

Solution:

Given: In trapezium ABCDand are mid-points of AB and DC, MN">AB and MN">DC.

To prove: AD = BC

Construction: Join CM and DM.

Proof:

In ΔCMN and ΔDMN,

MN = MN                           (Common sides)
">CNM = ">DNM = 90°">°°     (Given, MN">DC)
CN = DN                             (Given, N is the mid-point DC)

"> By SAS congruence criteria,
ΔCMN "> ΔDMN

So, CM = DM                  (CPCT)    .....(i)
And, ">CMN = ">DMN    (CPCT)
But, ">AMN = ">BMN = 90°">°°      (Given, MN">AB)
">">AMN -">- ">CMN = ">BMN -">- ">DMN
">">AMD = ">BMC           .....(ii)

Now, in ΔAMD and ΔBMC,

DM = CM                            [From (i)]
">AMD = ">BMC                [From (ii)]
AM = BM                            (Given, M is the mid-point AB)

"> By SAS congruence criteria,
ΔAMD "> ΔBMC

Hence, AD = BC    (CPCT)