jmc

Since $\angle BCA=40^{\circ }$ and $△ADC$ is isosceles with $AD=DC,$ we know $\angle DAC=\angle ACD=40^{\circ }.$

Since the sum of the angles in a triangle is $180^{\circ },$ we have $$\begin{array}{rl}\angle ADC& =180^{\circ }-\angle DAC-\angle ACD\\ & =180^{\circ }-40^{\circ }-40^{\circ }\\ & =100^{\circ }.\end{array}$$Since $\angle ADB$ and $\angle ADC$ are supplementary, we have $$\begin{array}{rl}\angle ADB& =180^{\circ }-\angle ADC\\ & =180^{\circ }-100^{\circ }\\ & =80^{\circ }.\end{array}$$Since $△ADB$ is isosceles with $AD=DB,$ we have $\angle BAD=\angle ABD.$ Thus, $$\begin{array}{rl}\angle BAD& =\frac{1}{2}(180^{\circ }-\angle ADB)\\ & =\frac{1}{2}(180^{\circ }-80^{\circ })\\ & =\frac{1}{2}(100^{\circ })\\ & =50^{\circ }.\end{array}$$Therefore, $$\begin{array}{rl}\angle BAC& =\angle BAD+\angle DAC\\ & =50^{\circ }+40^{\circ }\\ & =\boxed{90^{\circ }}.\end{array}$$