Question

Through the vertex D of a parallelogram ABCD, a line is drawn to intersect the sides BA and BC produced at E and F respectively. Prove that DA/AE= FB/BE= FC /CD

Answer 1

Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

•Similarity of triangles should be expressed symbolically using correct correspondence of their vertices
•AA similarity: if two angles of one triangle are respectively equal to two angles of another triangle then the two Triangles are similar.

SOLUTION:

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.

HOPE THIS WILL HELP YOU..

Answer 2

Hey...!!!
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ABCD is a parallelogram and E is a point on AB, DE is produced to meet CB produced
At F.

Consider ΔADE and ΔFBE
<AED = <BEF (vertical angles)
<ADE = <BFE (alternate angle)

⇒ ΔADE ~ ΔFBE (AA criterion)

⇒ AD/BF = AE/BE (Corresponding sides of similar triangles are proportional)

⇒ AD/AE = BF/BE --------------- (1)


Similarly, ΔFCD ~ ΔFBE  [DC // EB]
CD/BE = FC/BF
BF/BE = FC/CD --------------- (2)

From equations 1 and 2, we get
AD/AE = BF/BE = FC/CD.
Hence proved.

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I Hope it's help you....!!!!