Question

If a line is drawn parallel to one side of a triangle, to intersect the other two sides, the other two sides
are divided in the same ratio. Using above theorem solve the following:
a. In AABC, D and E are the points on AB and AC such that DE||BC AD = 2cm, AB = 6cm, AC
= 9cm find AE
b. In AABC, D and E are the points on AB and AC such that DE||BC AD = x, DB = x - 2, AE = X
+2 EC = x - 1 find x

please give full explanation answer ​

Answer 1

Answer:

Given:

DE∣∣BC

To prove that:

AE

EC

=

AD

BD

Proof:

∠AED=∠ACB Corresponding angles

∠ADE=∠ABC Corresponding angles

∠EAD is common to both the triangles

⇒ΔAED∼ΔACB by AAA similarity

⇒

AE

AC

=

AD

AB

⇒

AE

AE+EC

=

AD

AD+BD

⇒

AE

EC

=

AD

BD

Hence proved

solution

Answer 2

$\huge \tt{ \underline{\underline{\blue{Answer: - }}}}$

$\bold{ \underline{Given: - }}$

$\sf{DE∣∣BC}$

$\bold{ \underline{ To \: prove \: that:- }}$

$\sf{ \dfrac{EC}{AE}</p><p></p><p> </p><p> = \dfrac</p><p>{BD}{AD</p><p>}}$

$\bold{ \underline{ Proof:- }}$

$\sf{∠AED=∠ACB \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Corresponding \: angles}$

$\sf{∠ADE=∠ABC \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Corresponding \: angles}$

$\sf{</p><p></p><p>∠EAD \: is \: common \: to \: both \: the \: triangles</p><p>}$

$\sf{⇒ΔAED∼ΔACB \: by \: AAA similarity}$

$\sf{⇒ \dfrac</p><p>{AC}{AE}</p><p> </p><p> =\dfrac {AB}{</p><p>AD</p><p>}}$

$\sf{⇒ \dfrac</p><p>{AE+EC}{AE}</p><p></p><p> </p><p> = \dfrac{AD+BD</p><p>}{AD}}$

$\sf{</p><p>⇒ \sf \green{ \dfrac{EC}{AE}</p><p></p><p> </p><p> = \dfrac</p><p>{BD}{AD</p><p>}}</p><p> }$

$\boxed {\bold{\red{Hence \: proved \: }}}\:\:\:{ \bigstar}$

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$\huge \tt{ \underline{\underline{\blue{Answer: - }}}}$

$\implies{\sf{ \dfrac{EC}{AE}</p><p></p><p> </p><p> = \dfrac</p><p>{BD}{AD</p><p>}}}$

$\implies{\sf{ \dfrac{9}{AE}</p><p></p><p> </p><p> = \dfrac</p><p>{6}{2</p><p>}}}$

$\implies{ \sf{ \dfrac{18}{6} = AE}}$

$\implies{ \sf \green{ 3cm = AE}}$

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$\huge \tt{ \underline{\underline{\blue{Answer: - }}}}$

$\: \implies\sf{ \dfrac{EC}{AE} = \dfrac{BD}{AD}}$

$\: \implies\sf{ \dfrac{x-1}{x+2} = \dfrac{x-2}{x}}$

$\: \implies\sf{ (x-1)}{(x) = ({x-2)}{(x + 2)}}$

$\: \implies\sf{ ( {x}^{2} -x) = ({ {x}^{2} -{2}^{2} )}{}}$

$\: \implies\sf{ ( {x}^{2} - {x}^{2} - x ) = ({ -{2}^{2} )}{}}$

$\: \: \implies\sf{ ( \cancel{x}^{2} - \cancel{x}^{2} - x ) = ({ -{2}^{2} )}{}}$

$\: \: \implies\sf{ ( - x ) = ({ -{2}^{2} )}{}}$

$\: \: \implies\sf{ ( - x ) = ({ 4 )}{}}$

$\: \: \implies\sf \green{ x = { - 4 }{}}$

$\large\sf{\red{@\:Aꪀցꫀl}}$

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