ABC is a triangle. Extended CB to D to make the angle ABD = 3 times of the angle A. Find the angle A, angle C and the angle ABC.

Solution:

Let the angle A be x and the angle C be y

since AB = AC (Given), so the angle C = angle ABC = y

Principles used: If two sides of a triangle are equal, then the triangle is an isosceles triangle. In an isosceles triangle, its two base angles are equal.

Given: the angle ABD = 3 times the angle A

since extended CB to D, so the angle ABD is an exterior angle of the triangle ABC.

angle ABD = angle A + angle C (exterior angle of a triangle theorem)

so 3x = x + y (exterior theorem)

In triangle ABC, angle A + angle C + angle ABC = 180^{o}

x + y + y = 180^{o} (the sum of the interior angles in a triangle is 180^{o}

we get two equations:

3x = x + y ...equation1

x + y + y = 180^{o} ...equation2

from equation1: 2x = y

from equation2: x + 2y = 180^{o}

substitute y = 2x into the equation x + 2y = 180^{o}

x + 2(2x) = 180^{o}

x + 4x = 180^{o}

5x = 180^{o}

x = 180^{o}/5 = 36^{o}

y = 2x = 2(36^{o}) = 72^{o}

so the angle A is 36^{o}, angle C is 72^{o} and the angle ABC is also 72^{o}.