Question

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. Prove that AD = BC.

Answer 1

Given: In trapezium ABCD, M and N are mid-points of AB and DC, MN$\perp$AB and MN$\perp$DC.

To prove: AD = BC

Construction: Join CM and DM.

Proof:

In ΔCMN and ΔDMN,

MN = MN (Common sides)
$\angle$CNM = $\angle$DNM = 90$°$ (Given, MN$\perp$DC)
CN = DN (Given, N is the mid-point DC)

$\therefore$ By SAS congruence criteria,
ΔCMN $\cong$ ΔDMN

So, CM = DM (CPCT) .....(i)
And, $\angle$CMN = $\angle$DMN (CPCT)
But, $\angle$AMN = $\angle$BMN = 90$°$ (Given, MN$\perp$AB)
$\Rightarrow$$\angle$AMN $-$ $\angle$CMN = $\angle$BMN $-$ $\angle$DMN
$\Rightarrow$$\angle$AMD = $\angle$BMC .....(ii)

Now, in ΔAMD and ΔBMC,

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

$\therefore$ By SAS congruence criteria,
ΔAMD $\cong$ ΔBMC

Hence, AD = BC (CPCT)