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  • In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Q3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

   \frac{ar (ABC)}{ar ( DBC )} = \frac{AO}{DO}

   

Answers (1)

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Let DM and AP be perpendicular on BC.

area\,\,of\,\,triangle=\frac{1}{2}\times base\times perpendicular

\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}

In \triangle APO\, and\, \triangle DMO,

\angle APO=\angle DMO         (Each 90 \degree)

\angle AOP=\angle MOD         (Vertically opposite angles)

\triangle APO\, \sim \, \triangle DMO,       (AA similarity)

\frac{AP}{DM}=\frac{AO}{DO}    

Since 

 \frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}

\Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{AP}{ MD}=\frac{AO}{DO}

 

 

 

 

 

 

 

 

 

 

 

Posted by

seema garhwal

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