Question

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

ABCD is a trapezium with AB || CD. M and N are the mid-points of sides AB and AC, respectively.

Join AN and BN.

In ∆AMN and ∆BMN,

AM = BM (M is the mid-point of AB)

∠AMN = ∠BMN (MN ⊥ AB)

MN = MN (Common)

∴ ∆AMN ≅ ∆BMN (SAS congruence axiom)

So, AN = BN .....(1) (CPCT)

∠ANM = ∠BNM (CPCT)

Now,

∠DNM = ∠CNM (90º each)

∴ ∠DNM − ∠ANM = ∠CNM − ∠BNM

⇒ ∠AND = ∠BNC .....(2)

In ∆AND and ∆BNC,

DN = CN (N is the mid-point of CD)

∠AND = ∠BNC [From (2)]

AN = BN [From (1)]

∴ ∆AND ≅ ∆BNC (SAS congruence axiom)

So, AD = BC (CPCT)

Hence proved.