Feb 9, 2018

$\angle \left(B D C\right) = {64}^{0}$

#### Explanation:

Given $A \hat{B} C = A \hat{C} B = {32}^{0}$ as $A B = A C$

Consider isosceles triangle ABC,

$B \hat{A} C = 180 - \left(A \hat{B} C + A \hat{C} B\right) = 180 - 32 - 32 = {116}^{0}$

ABCD is a cyclic quadrilateral. Therefore $B \hat{A} C + B \hat{D} C = {180}^{0}$

$B \hat{D} C = 180 - B \hat{A} C = 180 - 116 = {64}^{0}$