Question

in traingle ABC BE parallel BC and AC=5.6cm AE= 2.1 CM find AD is to DB​

Answer 1

$\huge\underbrace\mathfrak\color{lime}{ †\: Answer:–}$

Given ;

ABC is a triangle .

DE || BC

AC = 5.6cm

AE = 2.1cm

To find ;

AD and DB

Now ;

Here DE || BC (given)

∴ ∠ADE = ∠ABC $~~~~~~~~~(parallel \:lines \:cut \:by \:transversal)$

∴ ∠AED = ∠ACB $~~~~~~~~~(parallel \:lines \:cut \:by \:transversal)$

Then ;

$~~~~~~~~~~~~~~~ In ∆ABC and ∆ADE$

$~~~~~~~~~~~~~~~⇒∠A = ∠A$ (common difference )

$~~~~~~~~~~~~~~~⇒∠ADE = ∠ABC$ (given)

$~~~~~~~~~~~~~~~⇒∠AED = ∠ACB$ (given)

By AAA congruence rule

$~~~~~~~~~~~~~~~∆ABC ≅ ∆ADE$

So now substitute the value ;

$~~~~~~~~~~~~~~~⇒\frac{AE}{AC} = \frac{AD}{AB} => \frac{2.1}{5.6}$

So DB is AB - AD

$~~~~~~~~~~~~~~~⇒DB = AB - AD$

$~~~~~~~~~~~~~~~⇒DB = 5.6 - 2.1$

$~~~~~~~~~~~~~~~⇒DB = 3.5$

So ;

$~~~~~~~~~~~~~~~⇒AD = 2.1$

$~~~~~~~~~~~~~~~⇒DB = 3.5$

Therefore ;

$~~~~~~~~~~~~~~~⇒\frac{AD}{DB} = \frac{2.1}{3.5}$

Answer 2

$\huge\underbrace\mathfrak\color{lime}{ †\: Answer:–}$

Given ;

ABC is a triangle .

DE || BC

AC = 5.6cm

AE = 2.1cm

To find ;

AD and DB

Now ;

Here DE || BC (given)

$∴ ∠ADE = ∠ABC ~~~~~~~~~(parallel \:lines \:cut \:by \:transversal)$

$∴ ∠AED = ∠ACB ~~~~~~~~~(parallel \:lines \:cut \:by \:transversal) \\ \\ </p><p>Then ;</p><p>~~~~~~~~~~~~~~~ In ∆ABC and ~~~~~~~~~~~~~~~⇒∠A = ∠A ⇒∠A=∠A (common difference )</p><p></p><p>~~~~~~~~~~~~~~~⇒∠ADE = ∠ABC ⇒∠ADE=∠ABC (given)</p><p></p><p>~~~~~~~~~~~~~~~⇒∠AED = ∠ACB ⇒∠AED=∠ACB (given)</p><p></p><p>By AAA congruence rule</p><p></p><p>~~~~~~~~~~~~~~~∆ABC ≅ ∆ADE ∆ABC≅∆ADE</p><p></p><p>So now substitute the value ;</p><p>~~~~~~~~~~~~~~~⇒\frac{AE}{AC} = \frac{AD}{AB} = > \frac{2.1}{5.6} ⇒ </p><p>AC</p><p>AE</p><p> </p><p> = </p><p>AB</p><p>AD</p><p> </p><p> => </p><p>5.6</p><p>2.1</p><p> </p><p> </p><p></p><p>So DB is AB - AD</p><p></p><p>~~~~~~~~~~~~~~~⇒DB = AB - AD ⇒DB=AB−AD</p><p></p><p>~~~~~~~~~~~~~~~⇒DB = 5.6 - 2.1 ⇒DB=5.6−2.1</p><p></p><p>~~~~~~~~~~~~~~~⇒DB = 3.5 ⇒DB=3.5</p><p></p><p>So ;</p><p>~~~~~~~~~~~~~~~⇒AD = 2.1 ⇒AD=2.1</p><p></p><p>~~~~~~~~~~~~~~~⇒DB = 3.5 ⇒DB=3.5</p><p></p><p>Therefore ;</p><p>~~~~~~~~~~~~~~~⇒\frac{AD}{DB} = \frac{2.1}{3.5} ⇒ </p><p>DB</p><p>AD</p><p> </p><p> = </p><p>3.5</p><p>2.1</p><p>$