Question

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

$\text{Prove that :}\left(i\right)\Delta DMC=\Delta BNAandsoCM=AN\left(ii\right)\Delta AMD\cong \Delta CNBandsoAM=CN\left(iii\right)\text{ANCM is a parallelogram.}$

Answer 1

$\text{Given : In parallelogram ABCD, poiints M and N lie on the diagonal BD such that}DM=BN\text{AN, NC, CM and MA are joined}\text{To prove :}\left(i\right)\Delta DMC\cong \Delta BNAandsoCM=AN\left(ii\right)\Delta AMD\cong \Delta CNBandsoAM=CN\left(iii\right)\text{ANCM is a parallelogram}\text{Proof :}\left(i\right)\text{In DMC and}\Delta BNA.CD=AB\left(\text{opposite sides of}||gmABCD\right)DM=BN\left(\text{given}\right)\angle CDM=\angle ABN\left(\text{alternate angles}\right)\therefore \Delta DMC\cong \Delta BNA\left(\text{SAS axiom}\right)\therefore CM=AN\left(\text{c.p.c.t.}\right)Similarly,in\Delta AMDand\Delta CNBAD=BC\left(\text{opposite sides of}||gm\right)DM=BN\left(\text{given}\right)\angle ADM=\angle CBN\left(\text{alternate angles}\right)\therefore \Delta AMD\cong \Delta CNB\left(\text{SAS axiom}\right)\therefore AM=CN\left(\text{c.p.c.t.}\right)\left(iii\right)\because CM=ANandAM=CN\left(\text{proved}\right)\therefore \text{ANCM is a parallelogram}\left(\text{opposite sides are equal}\right)\text{Hence proved.}$