plz help me!

Asked by  | 10th Mar, 2009, 12:33: PM

Expert Answer:

Note that,

In triangles DEA and FEB,

angle DEA=angle FEB(vertically opposite angles)

angle EDA=angle EFB( alternate int angles , with DA parallel to CB i.e. CF)

So,

 triangle DEA similar to triangle FEB (AA rule)

DA/EA=FB/EB...(i)

Next,

 in triangles FEB and FDC,

angle F is common

angle FEB = angle FDC(as EB and DC are parallel, so corr angles are equal)

so,

 triangle FEB similar to triangle FCD.

sO,

 FB/EB=FC/DC...(II)

(I) AND (II) taken together prove the result..

 

 

 

 

 

Answered by  | 13th Mar, 2009, 03:20: PM

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