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.
Given : ABCD is trapezium
M, N are mid-point of AB and DC respectively and perpendicular to both AB and DC
To prove : AD = BC
Proof: ANM and BNM
AM = BM (M is mid-point)
AMN = BMN = 90° (given)
MN = MN (common)
ANM BNM (congruency of SAS)
AN = BN (by CPCT) …(i)
We know that
ANM = BNM (by CPCT)
Then 90° – ANM = 90° – BNM
AND = BNC …(ii)
In AND and BNC,
AN = BN (From i)
AND = BNC (From ii)
DN = CN (N is mid-point)
AND BNC (by SAS congruency)
AD = BC (by CPCT)
Hence, proved.