Question

solve question no 12 and 14

Answer 1

Ques 12

In ΔDEC and ΔBEF,

∠DEC = ∠BEF (Vertically opposite angles)

CE = EB (E is the mid-point of CB)

∠DCE = ∠EBF (Alternate anlges ; AB II DC ; AB II AF)

∴ ΔDEC ≅ ΔBEF

BF = DC (By C.PC.T)

DC = AB (Opposite sides are equal in parallelogram)

BF  = AB

AF = AB + BF = AB + AB = 2AB

Hence Proved.

Ques 14

In quadrilateral AMCN,

AM = CN (M and N are the midpoints of equal sides AB and CD)

AM II CN (AB II CD)

∴AMCN is a parallelogram.

∴AM II NC and AM = NC (Opposite sides are equal and parallel in a parallelogram)

In quadrilateral PMQN,

PN II MQ (AN II NC)

PN = MQ ( P and Q are mid-points of equal sides AN and MC)

∴ PMQN is a parallelogram.

Hence, Proved...

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