Answer:
[tex] m\angle BDC = 95\degree [/tex]
Step-by-step explanation:
[tex] In\: \triangle ACD\: \&\:\triangle BDC[/tex]
[tex] m\angle A = m\angle B[/tex] (given)
[tex]\because m\angle A = 50\degree [/tex]
[tex]\therefore m\angle B = 50\degree [/tex]
[tex] m\angle ADC = m\angle BCD[/tex] (given)
[tex]\because m\angle ADC = 35\degree [/tex]
[tex]\therefore m\angle BCD= 35\degree [/tex]
[tex] In\: \triangle BDC[/tex]
[tex] m\angle BDC = 180\degree - (m\angle B+m\angle BCD) [/tex]
(By angle sum postulate of a triangle)
[tex] m\angle BDC = 180\degree - (50\degree +35\degree ) [/tex]
[tex] m\angle BDC = 180\degree - (50\degree +35\degree ) [/tex]
[tex] m\angle BDC = 180\degree - 85\degree [/tex]
[tex] m\angle BDC = 95\degree [/tex]
