Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • 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)

best_answer

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

View full answer

Similar Questions

Trending Articles/News

anam-khan

CBSE allows Class 10 students with basic math to study mathematics in 11th for 2025-26
anam-khan

CBSE announces post-result schedule for Class 10, 12 board exams 2025; apply for answer sheet from May 21

Latest Question