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  • 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.

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.

Answers (1)

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: \triangleANM and \triangleBNM

AM = BM                                           (M is mid-point)

\angleAMN = \angleBMN = 90°          (given)

MN = MN                                           (common)

\triangleANM \cong \triangleBNM                     (congruency of SAS)

AN = BN                                             (by CPCT)                  …(i)

We know that

\angleANM = \angleBNM                    (by CPCT)                 

Then 90° – \angleANM = 90° – \angleBNM

\Rightarrow \angleAND = \angleBNC                                                                …(ii)

In \triangleAND and \triangleBNC,

AN = BN                                             (From i)

\angleAND = \angleBNC                                 (From ii)

DN = CN                                             (N is mid-point)

\triangleAND \cong \triangleBNC                                   (by SAS congruency)

AD = BC                                             (by CPCT)

Hence, proved.

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