Question

Consider the parallelogram ABCD as shown in the figure, where AE/AB=CF/CD =1/n for some positive integer n.


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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

In ∆ EAD & ∆DCF,

∠1 = ∠ 2  [ corresponding angles are equal , as AB || DC]

∠3 = ∠ 4  [ corresponding angles are equal , as AD || BC]

∆ EAD ~ ∆DCF

[By AA similarity criterion]

EA/DC = AD/CF = DE/FD

[corresponding sides of similar triangles are proportional]

AD/AE = CF/CD…………….(1)

In ∆ EAD & ∆EBF,

∠1 = ∠ 1     (Common angle)

∠3 = ∠ 4      [ corresponding angles are equal , as AD || BC]

∆ EAD  ~ ∆EBF

[By AA similarity criterion]

EA/EB = AD/BF = DE/EF

AD/AE= FB/BE……………….(2)

From eq 1 & 2

AD/AE= FB/BE = CF/CD

Hence, proved.

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