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in Triangles by (34.5k points)
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In a trapezium ABCD, AB || DC, and also EF || AB, then AE/ED =

(A) AB/DC

(B) AB/EF

(C) BF/AE

(D) BF/FC

2 Answers

+1 vote
by (57.0k points)
selected by
 
Best answer

Correct option is (D) BF/FC

Given AB || DC & EF || AB

\(\therefore\) DC || EF

Construction :

Join BD which intersect EF at G.

In \(\triangle ABD,\) EG || AB

\(\therefore\) \(\triangle EDG\sim\triangle ADB\)       (Corresponding angles are equal as EG || AB)

\(\therefore\) \(\frac{ED}{AD}=\frac{GD}{BD}\)          (Properties of similar triangles)

\(\therefore\) \(\frac{ED}{AD-ED}=\frac{GD}{BD-GD}\)   \((\because\,If\,\frac ab=\frac cd\Rightarrow\frac a{b-a}=\frac c{d-c})\)

\(\Rightarrow\) \(\frac{ED}{AE}=\frac{GD}{BG}\)

\(\Rightarrow\) \(\frac{AE}{ED}=\frac{BG}{GD}\)       ________________(1)

In \(\triangle BCD,\) GF || DC

\(\therefore\) \(\triangle BGF\sim\triangle BDC\)

\(\therefore\) \(\frac{BG}{BD}=\frac{BF}{BC}\)

\(\Rightarrow\) \(\frac{BG}{BD-BG}=\frac{BF}{BC-BF}\)

\(\Rightarrow\) \(\frac{BG}{GD}=\frac{BF}{FC}\)       ________________(2)

From (1) & (2), we obtain

\(\frac{AE}{ED}\) = \(\frac{BF}{FC}\)

+1 vote
by (35.6k points)

Correct option is: (D) \(\frac{BF}{FC}\)

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