D is any point on side QR of PQR .A line drawn through D parallel to PR meets PQ in F and a line drawn through D parallel to PQ meets PR in E. show that ar(∆EQD)= ar (∆FDR).
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In parallelogram EFDQ and in parallelogram FEDR
area of triangle QED =1/2 of area of parallelogram EFDQ----(1)
area of triangle FDR=1/2 of area of parallelogram FEDR------(2)
DQ=DR as D is the mid point of Q and R
from (1)and (2)
area of triangle QED=area of triangle FDR
area of triangle QED =1/2 of area of parallelogram EFDQ----(1)
area of triangle FDR=1/2 of area of parallelogram FEDR------(2)
DQ=DR as D is the mid point of Q and R
from (1)and (2)
area of triangle QED=area of triangle FDR
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