Solve if you know (no spam)
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Answer:
Yes
Step-by-step explanation:
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)
EL
BL
=
AE
BC
[By CPCT]
EL
BL
=
2BC
BC
EL
BL
=
2
1
EL=2BL
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