In an acute-angled triangle ABC, the height BH and the bisector AD are drawn, intersecting

In an acute-angled triangle ABC, the height BH and the bisector AD are drawn, intersecting at point O. It turned out that the angle AOB is four times the angle DAB. What is the CAB angle?

Let’s denote the value of the angle DAB by x. Since AD is a bisector, then <CAD = <DAB = x or <OAC = <BAO = x.

Triangle ABH is rectangular because the height BH is perpendicular to the AC.

The sum of the angles of triangle ABH:

<DAB + <CAD + <ABH + <AHB = 180 °

x + x + <ABH + 90 ° = 180 °;

2x + <AHB = 90 °

<AHB = 90 ° – 2x.

Find the sum of the angles of the triangle AOB, taking into account the condition – <AOB = 4 * <DAB = 4x:

<DAB + <AOB + <AHB = 180 °;

x + 4x + (90 ° – 2x) = 180 °;

3x = 90 °;

x = 30 °;

<CAB = <CAD + <DAB = x + x = 30 ° + 30 ° = 60 °.