Question

In triangle ABC,she DE paralel side BC AD=5 ,DB=10, AE=4 find EC​

Answer 1

$\huge\underline\mathtt{♡Be Brainly♡}$

Answer 2

$\huge{\underline{\overline{\mathfrak{\red{Solution}}}}}$

$\underline\mathtt\blue{Given:}$

$\sf{DE \: ll \: BC}$

AD=5cm ,DB=10cm ,AE=4cm

$\underline\mathtt\blue{To\: find:}$

$\sf{EC=?}$

$\rule{300}{2}$

$\sf{Now,}$

In triangle ABC,

DE ll BC [Given]

Hence,

$\sf{\frac{AD}{DB} = \frac{AE}{EC} }$

From, Thales Theorem ,

$\sf{\implies \frac{DB}{EC} = \frac{AD}{AE} }$

Now,by putting the values,we get,

$\sf{\implies \frac{10}{EC} = \frac{5}{4} }$

$\sf{\implies EC = \frac{10 \times 4}{5} }$

$\sf{\implies EC=\cancel\frac{40}{5}}$

$\sf{\implies EC \: = \: 8\: cm}$