Question

In the diagram, AD=CD=CB and measure of angle A=40°. How many degrees are in angle DCB?​

Answer 1

Answer:

$20^{\circ}$

Step-by-step explanation:

Consider triangle ACD. This triangle is isosceles triangle, because AD = DC. Angles adjacent to the base AC are congruent abgles, so

$m\angle DAC=m\angle DCA=40^{\circ}$

The sum of the measures of all interiror angles in the triangle is always $180^{\circ},$ then

$m\angle DAC+m\angle DCA+m\angle ADC=180^{\circ}\\ \\40^{\circ}+40^{\circ}+m\angle ADC=180^{\circ}\\ \\m\angle ADC=180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ}$

Angles ADC and BDC are supplementary angles (add up to $180^{\circ}$), then

$m\angle BDC=180^{\circ}-100^{\circ}=80^{\circ}$

Consider triangle BCD. This triangle is isosceles triangle because BC = DC. Angles adjacent to the base BD are congruent abgles, so

$m\angle BDC=m\angle DBC=80^{\circ}$

The sum of the measures of all interiror angles in the triangle is always $180^{\circ},$ then

$m\angle DBC+m\angle BDC+m\angle BCD=180^{\circ}\\ \\80^{\circ}+80^{\circ}+m\angle BCD=180^{\circ}\\ \\m\angle BCD=180^{\circ}-80^{\circ}-80^{\circ}=20^{\circ}$