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The median BE and CF of ABC intersects at G. Prove that ar(GBC) = ar(quad AFGE)

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert
Answer:

Join EF

EF will be parallel to BC.(mid point theorem)

ar (BEF) = ar(CEF) ( as area of triangles on same base and same two parallel lines are equal)

ar(BEF) - ar(GEF) = ar CEF - arGEF

arBFG = arCGE …(i)

Now ar (BEC) = ar(ABE) (BE is median)…(ii)

ar (BEC)- arCGE= ar(ABE)- arBFG equal areas subtracted from both sides in eq(ii)

Ar(BGC) = ar(quad AFGE)

Answered by Expert 4th June 2014, 3:23 PM
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