Question

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

Prove that :
(i) ZDAB = ZCBA
(ii) ZADC = ZBCD
(iii) AC = BD
(iv) OA = OB and OC = OD

Answer 1

Answer:

(i) In ∆ABC and ∆BAD

AD=BD (GIVEN)_ _ _(1)

AB=AB (GIVEN)_ _ _(2)

ar∆ABC=ar∆BAD ( that is,∆ABC and ∆BAD are between same parrellel and have same base)

So,AC=BD_ _ _(3)

From (1),(2)&(3)

∆ABC=∆BAD (By SSS rule)

So, DAB=CBA.

(ii). Similarly from answer (i) ,we can prove the answer of (ii).

So,ADC=BCD.

(iii). In ∆ACD and ∆BDC

AD=BC (GIVEN)

DC=DC (GIVEN)

ar∆ACD=ar∆BDC (that is,∆ACD and ∆BDC are between same parrellel and have same base)

So,AC=BD.

(iv). In ∆AOD and ∆BOC

AD=BC (GIVEN)

AOD=BOC (That is, vertically opposite angles)

ADB=ACB (That is,∆ABC=∆BAD,proved)

CAD=CBD (That is,∆ACD=∆BCD,proved)

By ASA congruent rule

∆AOD=∆BOC

So,OA=OB and OC=OD.