The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC.

Prove that :

(i) ∠DAB = ∠CBA

(ii) ∠ADC = ∠BCD

(iii) AC = BD

(iv) OA = OB and OC = OD

Draw DM ⊥ AB and CN ⊥ AB.

(i) In △ DAM and △ CBN,

⇒ AD = BC (Given)

⇒ ∠DMA = ∠CNB (Both equal to 90°)

⇒ DM = CB (Since, DC || AB, DM ⊥ AB and CN ⊥ AB)

∴ △ DAM ≅ △ CBN (By R.H.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ ∠DAM = ∠CBN ……..(1)

From figure,

⇒ ∠DAM = ∠DAB

⇒ ∠CBN = ∠CBA

Substituting above values in equation (1), we get :

⇒ ∠DAB = ∠CBA.

Hence, proved that ∠DAB = ∠CBA.

(ii) Since, △ DAM ≅ △ CBN

∴ ∠ADM = ∠BCN (By C.P.C.T.C.)

⇒ ∠ADM + 90° = ∠BCN + 90°

⇒ ∠ADC = ∠BCD.

Hence, proved that ∠ADC = ∠BCD.

(iii) In △ ADB and △ BCA,

⇒ AB = AB (Common side)

⇒ ∠DAB = ∠CBA (Proved above)

⇒ AD = BC (Given)

∴ △ ADB ≅ △ BCA (By S.A.S. axiom)

⇒ AC = BD (By C.P.C.T.C.)

Hence, proved that AC = BD.

(iv) In △ OAD and △ OBC,

⇒ ∠OAD = ∠OCB (Alternate angles are equal)

⇒ ∠AOD = ∠BOC (Vertically opposite angles are equal)

⇒ AD = BC (Given)

∴ △ OAD ≅ △ OBC (By A.A.S. axiom)

⇒ OA = OB and OC = OD. (By C.P.C.T.C.)

Hence, proved that OA = OB and OC = OD.