Isosceles triangle ABC, D is the interior point, AB=BC, angle ABC = 80, angle DAC = 30, angle DCA = 40. Find the angle ADB extends the intersection of CD AB to E and AD BC to F, so that AG is perpendicular to BF to G because AB=BC and angle ABC=80 degrees, so the angle BCA= 50 degrees because angle DCA=40 degrees. Angle DAC=30 degrees so angle CEA=90 degrees, angle EDA= angle DCA+ angle DAC=70 degrees because angle BCA= angle BAC=50 degrees, angle DCA=40 degrees, angle DAC=30 degrees so angle BCE= angle BCA- angle DCA= 10 degree, angle BAF=20 degrees because angle ABC = Angle BFA= 180-80-20=80 degrees, so angle ABC= angle BFA, so AB=AF Because AG is BC=BA, angle BCE= angle package = 10 degrees, so triangle BCE is all equal to triangle package, so BE=BG= 1/2BF. The following is the process of hypothesis and verification: If BD=BF is to make the hypothesis hold, then the triangle BDE is a right triangle, that is, the angle BDE=BE/BD is 20 degrees because BD=BF and BFA=80 degrees, and the angle CBD=20 degrees because BCE= 10 degrees, so the angle BDE= angle CBD+ angle BCE=30 degrees. So the Sin angle BDE=65438 Because the angle EDA=70 degrees, the angle ADB= EDA+ BDE=70+30= 100 degrees. In triangle ABC, ∠ BAC = 45 degrees, and AD ⊥ BC is at point D BD.