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

`36^\circ, 108^\circ, 36^\circ`

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`