Question

In $\Delta$ ABC, D is a point on BC such that AB= AD = BD =DC. Show that : $\angle$ ADC : $\angle$ C = 4: 1.

Answer 1

Here AB,AD and BD are side of a triangle and all the sides are equal as per question.
This means that $\Delta$ is equilateral triangle and each angle is of

${60}^0$ .
Now $\angle ADC+\angle ADB={180}^0$

or $\angle ADB={60}^0$

as the $\Delta$ is equilateral triangle
$\angle ADC={180}^0-{60}^0={120}^0$ ........1
Now in $\Delta$ADC ,

AD= DC{ given }

So the $\Delta$ is isosceles triangle .

i.e $\angle DAC=\angle ACD$.........2
Let the angle be $y^0$

Now we know that sum of all the angle of triangle is ${180}^0$.

or,
$\angle ACD+\angle ADC+\angle DAC={180}^0$

or,$\angle ADC+2y={180}^0$

or,${120}^0+2y={180}^0$
{by eq 1 and 2}

⇒2y= ${60}^0$

⇒y=${30}^0$
Now $\angle ADC$

$\angle C=\frac{{120}^0}{{30}^0}$

= $\frac{4}{1}$

Hence proved