Question

In △ABC point D is on side BC such that AD = BD and AB = DC = AC. Find all the angles of △ABC.

Answer 1

Answer:

$36^\circ, 108^\circ, 36^\circ$

Step-by-step explanation:

In $\triangle ABC$, sides AB = AC.

We know the property that angles opposite to equal sides in a triangle are equal.

Hence,  $\angle ABC = \angle ACB$

Let this angle be x.

So, $\angle ABC = \angle ACB = x ...... (1)$

Similarly, in $\triangle ABD$

Hence, $\angle ABD = \angle BAD$

$\angle ABD$ and $\angle ABC$ are same.

By equation (1):

$\angle ABD = \angle BAD = x ...... (2)$

Similarly, in $\triangle ADC$:

$\angle ADC = \angle DAC$

Let this angle be y.

$\Rightarrow \angle ADC = \angle DAC = y ...... (3)$

We know that sum of all three angles in a triangle is equal to $180 ^\circ$.

In $\triangle ADC$, sum of all three angles:

$x + y + y = 180^\circ\\\Rightarrow x + 2y = 180 ...... (4)$

In $\triangle ABC$, sum of all three angles:

$x + (x+y) + x = 180\\\Rightarrow 3x + y = 180 ...... (5)$

Using elimination method to solve equation (4) and (5):

Multiplying equation (5) by 2 and subtracting (4) from it:

$5x = 180\\\Rightarrow x = 36^\circ$

Putting value of x in (4):

$36^\circ + 2y = 180\\\Rightarrow y = 72^\circ$

So, angles of $\triangle ABC$ are:

x, (x+y) and y

$\Rightarrow 36^\circ, 108^\circ \text{ and } 36^\circ$