Answer:
[tex]20^{\circ}[/tex]
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
[tex]m\angle DAC=m\angle DCA=40^{\circ}[/tex]
The sum of the measures of all interiror angles in the triangle is always [tex]180^{\circ},[/tex] then
[tex]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}[/tex]
Angles ADC and BDC are supplementary angles (add up to [tex]180^{\circ}[/tex]), then
[tex]m\angle BDC=180^{\circ}-100^{\circ}=80^{\circ}[/tex]
Consider triangle BCD. This triangle is isosceles triangle because BC = DC. Angles adjacent to the base BD are congruent abgles, so
[tex]m\angle BDC=m\angle DBC=80^{\circ}[/tex]
The sum of the measures of all interiror angles in the triangle is always [tex]180^{\circ},[/tex] then
[tex]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}[/tex]
