A quadrilateral ABCD is inscribed in a circle. Beams AB and DC meet at point K, and diagonals AC and BD

A quadrilateral ABCD is inscribed in a circle. Beams AB and DC meet at point K, and diagonals AC and BD meet at point N. The BNC angle is 68 and the AKD angle is 36. Find the angle BAC.

Since the angle between the intersecting chords is equal to the half-sum of the arcs on which this angle and the opposite one rests, then the angle BNC = (BC + AD) / 2.

Let us express the degree measure of the arc AD. Arc AD = 2 * 68 – BC = (136 – BC).

The angle between the secants drawn from one point is equal to the half-difference of the degree measures of the arcs formed by the points of intersection with the circle.

Angle AKD = (AD – BC) / 2.

Let us express the degree measure of the arc AD. Arc AD = 2 * 36 + BC = (72 + BC).

Then: (136 – BC) = (72 + BC).

2 * BC = 136 – 72 = 64.

BC = 64/2 = 32.

Then the inscribed angle BAC is equal to half of the degree measure of the BC arc.

Angle BAC = 32/2 = 16.

Answer: The BAC angle is 16.