Question

in the fig ab=cb,ab=cd and ef bisects bd at g prove that g is midpointof ef

Answer 1

AB=CB      eq.1

AB=CD     eq.2

from eq.1 and 2 we know that AB=CB=CD

therefore ABCD is a square.

EF bisects BD at G.

therefore EG=FG

consider triangle FGB and EGD.

EG=FG

angleDEG= angleGFB (alternate interior angles)(AB is parallel to CD)

angleDGE=angleBGF (vertically opposite angles)

therefore by ASA congruency both triangles are congruent

BG=DG (cpct)

hence proved

Answer 2

Answer:

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