Answer:
[tex]\angle A = 60^{\circ}[/tex] , [tex]\angle B = 80^{\circ}[/tex]and [tex]\angle C = 40^{\circ}[/tex]
Step-by-step explanation:
We are given that BD = DC
BD= DC
So, [tex]\angle DBC = \angle BCD[/tex] (Opposite angles of equal sides are equal) ----1
We are given that ∠BDC = 100°
In ΔBDC
Angle sum property of triangle : The sum of all angles of triangle is 180°
So, [tex]\angle BDC+\angle DBC +\angle BCD = 180^{\circ}[/tex]
[tex]100^{\circ}+2 \angle DBC= 180^{\circ}[/tex](Using 1)
[tex]2 \angle DBC = 180^{\circ}-100^{\circ}[/tex]
[tex]2 \angle DBC = 80^{\circ}[/tex]
[tex]\angle DBC = 40^{\circ}[/tex]
[tex]\angle DBC = \angle BCD=\angle 2 = 40^{\circ}[/tex]
So, [tex]\angle C = 40^{\circ}[/tex]
We are given that [tex]\angle 1 = \angle 2[/tex]
So,[tex]\angle 1 = \angle 2 = 40^{\circ}[/tex]
Now[tex]\angle BDC+\angle BDA = 180^{\circ}[/tex](Linear pair)
[tex]100^{\circ}+\angle BDA = 180^{\circ}\\\angle BDA =80^{\circ}[/tex]
In ΔABD
[tex]\angle ABD+\angle BDA+\angle BAD = 180^{\circ}[/tex](Angle sum property)
[tex]\angle 1 +\angle BDA+\angle BAD = 180^{\circ}\\40^{\circ}+80^{\circ}+\angle BAD = 180^{\circ}\\120^{\circ}+\angle BAD = 180^{\circ}\\\angle BAD =60^{\circ}[/tex]
So, [tex]\angle A =60^{\circ}\\\angle B = \angle 1+\angle 2 = 40^{\circ}+40^{\circ}=80^{\circ}[/tex]
Hence [tex]\angle A = 60^{\circ}[/tex] , [tex]\angle B = 80^{\circ}[/tex]and [tex]\angle C = 40^{\circ}[/tex]
