In △′sBMC and EMD, we have
∠BMC=∠EMD..........(opposite angles)
MC=MD........ (mid point)
∠MCB=∠MDE..........(alternate angles)
So, by AAS-congruence criterion, we have
△BMC≅△EMD
⇒ BC=ED [∵ Corresponding parts of congruent triangle a equal]
In △′sAEL and CBL, we have
∠ALE=∠CLB [Verticaly opposite angles]
∠EAL=∠BCL [Alternate angles]
So, by AA-criterion of similarity, we have
△AEL∼△CBL
⇒ AEBC=ELBL=ALCL
⇒ ELBL=AEBC
⇒ ELBL=AD+DEBC=BC+DEBC=2BCBC=2 [∵ AD=BC and DE=BC]
⇒ EL=2BL [Hence proved]
![1031848_1009347_ans_e2024f46d16742949133a78bdd83964b.png](https://search-static.byjusweb.com/question-images/toppr_ext/questions/1031848_1009347_ans_e2024f46d16742949133a78bdd83964b.png)