Altitudes \(\overline{AD}\) and \(\overline{BE}\) of acute triangle $ABC$ intersect at point $H$. If \(\angle AHB = 123^\circ\) and \( \angle BAH = 26^\circ\), then what is \(\angle HCA\) in degrees?
∠AHB = 123°
∠BAH = 26°
∠ABH = 31°
∠ABD = 64°
∠HBD = 33°
∠BHD = 57°
∠HAE = 33°
∠AHE = 57°
∠ACD = 57°
to be continued...
Why is this???
∠ABH = ∠HCA = 31°
Why is this???
