AP ICET 2017 Shift 2 Paper

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Question : 136 of 200

Marks: +1, -0

In ΔABC the bisector of the angle A meets BC in D and DE is parallel to BA meeting AC in E. then

{AE}/{EC}=

${AB}/{AC}$
${AC}/{AB}$
${BC}/{AB}$
${AB/{BC}$

Solution:

Using angle bisector theorem :

{AB}/{AC}={BD}/{CD}

....(i)

{AE}/{EC}={BD}/{CD}

...(ii)

[DE||BA]

From (i) and (ii) we get

{\AE}/{\EC}={\AB}/{\AC}

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