in the given figure Al is parallel to dc and E is the mid point of bc so that triangle Ebl is congruent to triangle Ecd
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proved that Δ EBL is congruent to ΔECD
Step-by-step explanation:
Δ EBL & ΔECD
∠BEL = ∠CED ( oppsoite angles)
∠EBL = ∠ECD ( as AL ║ CD)
∠ELB = ∠EDC ( as AL ║ CD)
BE = CE ( as E is mispoint of BC)
=> Δ EBL ≅ ΔECD
QED
Proved
Hence proved that Δ EBL is congruent to ΔECD
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