Question

in triangle ABC, DE parallel to BC and AC equal to 5.6 cm and AE equal to 2.1 CM then find AD : DB

Answer 1

hey mate.
here's the solution...

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 $~~~~~~~~~(parallel \:lines \:cut \:by \:transversal)$

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 ;

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