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In the figure ABCD is a …

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In the figure ABCD is a parallelogram and E divides BC in the ratio 1:3. BD and AE intersect at F. Show that DF = 4FB and AE = 4FE .

Posted by Nandini Saxena 8 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Lenin Valentine 5 years ago

AE = a +1/4d AF = ka + 1/4 kd AF = AD + DF = AD + hDB = d + h(a-d) = ha + (1-h)d AF = AF So, ka + 1/4kd = ha + (1-h)d From which k = h And 1/4k = 1-h since h=k 1/4k =1-k Giving k = 4/5 and therefore, h = 4/5 DF = hDB DF = 4/5DB DF : FB = 4:1 ∴ DF = 4FB Also AF = kAE AF : FE = 4:1 ∴ AF = 4FE

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