Question

The given figure shows parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN.

Prove that:
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM CN
(iii) ANCM is a parallelogram.

Answer 1

Given: In parallelogram ABCD, points M and N lie on the diagonal BD such that DM = BN
AN, NC, CM and MA are joined

To prove :
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM = CN
(iii) ANCM is a parallelogram

Proof :
(i) In ∆DMC and ∆BNA.

CD = AB (opposite sides of ||gm ABCD)

DM = BN (given)

∠CDM = ∠ABN (alternate angles)

∆DMC = ∆BNA (SAS axiom)

CM =AN (c.p.c.t.)

Similarly, in ∆AMD and ∆CNB

AD = BC (opposite sides of ||gm)

DM = BN (given)

∠ADM = ∠CBN – (alternate angles)

∆AMD = ∆CNB (SAS axiom)

AM = CN (c.p.c.t.)

(iii) CM = AN and AM = CN (proved)

ANCM is a parallelogram (opposite sides are equal)

Hence proved.