Question

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Answer 1

In △BMC and  △EMD

∠BMC=∠EMD  [Vertically opposite]

MC=DM [Given]

∠BCM=∠EDM [Alternate angles]

∴△BMC≅△EMD [By ASA]

Hence, BC=DE [By CPCT] →(1)

AE=AD+DE=BC+BC=2BC →(2)

Now,  △BLC∼△ELA (AA Similarly)

ELBL=AEBC   [By CPCT]

ELBL=2BCBC

ELBL=21

EL=2BL

Answer 2

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