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Question
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD.
Prove that :
(i) ΔABD and ΔECD are congruent.
(ii) AB = CE.
(iii) AB is parallel to EC
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Solution
Given: A ΔABC in which D is the mid-point of BC
AD is produced to E so that DE=AD
We need to prove that :
(i) ΔABD and ΔECD are congruent.
(ii) AB = CE.
(iii) AB is parallel to EC
(i) In ΔABD and ΔECD
BD = DC ...[ D is the midpoint of BC ]
∠ADB =∠CDE ...[ vertically opposite angles ]
AD = DE ...[ Given ]
∴ By Side-Angle-Side criterion of congruence, we have,
ΔABD ≅ ΔECD
(ii) The corresponding parts of the congruent triangles are congruent.
∴ AB = EC ...[ c.p.c.t .c]
(iii) Also, ∠BAD = ∠DEC ....[ c.p.c t.c ]
∠ABD = ∠DCE .....[ c.p.c t.c ]
AB || EC .....[ DAB and DEC are alternate angles ]
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