Mathematics
In the given figure, ABCD is a parallelogram, E is the mid-point of BC. DE produced meets AB produced at L. Prove that:
(i) AB = BL
(ii) AL = 2DC
Triangles
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Answer
(i) Given,
ABCD is a parallelogram.
We know that,
Opposite sides of a parallelogram are equal.
⇒ AB = CD and AD = BC
In △DEC and △BEL,
⇒ ∠LBE = ∠DCE (Alternate angles, since AB || DC)
⇒ EC = EB (Given)
⇒ ∠DEC = ∠BEL (Vertically opposite angles are equal)
∴ △DEC ≅ △BEL (By A.S.A axiom)
⇒ DC = BL (Corresponding parts of congruent triangles are equal)
Since, AB = DC
∴ AB = BL.
Hence, proved that AB = BL.
(ii) From figure,
⇒ AL = AB + BL
⇒ AL = DC + DC
⇒ AL = 2DC.
Hence, proved that AL = 2DC.
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