+1347 In triangle $ABC,$ altitudes $\overline{AD}$ and $\overline{BE}$ intersect at $H$. If $\angle BAC = 40^\circ$ and $\angle ABC = 40^\circ$, then what is $\angle AHB$?
+128707 Angle C = 180 - 40 - 40 = 100°
Angles ADC and BEC = 90° (since AD and BE are altitudes)
We have quadrilateral CEHD where the sum of its interior angles = 360
So angle EHD = 360 - 90 - 90 - 100 = 80°
And angle AHB is a vertical angle to angle EHD so it also measures 80°
