Geometry question on chord of the circle. link to the figure https://gyazo.com/ee38260f5ad7c52c48cbcf235b8e1c3c?

Let the chords intersect at point K. The vertical angles ∠AFC and ∠BKD are each equal to the mean of the arc measures of the intercepted arcs, and they are both right angles.

[arc(APC) + arc(BQD)]/2 = 90°
arc(APC) + arc(BQD) = 180°

The arcs have the same radius, and their sum is 180°. They therefore can be joined to form a circle.

Let O be the center of the circle.
Join BC, OA, OB, OC and OD. (Construction)

AB⊥CD (given)
∠ABC + ∠BCD + 90° = 180° (∠ sum of Δ)
Hence, ∠ABC + ∠BCD = 90°

∠AOC = 2 ∠ABC (∠ at center twice ∠ in segment)
∠BOD = 2 ∠BCD (∠ at center twice ∠ in segment)

∠AOC + ∠BOD = 2 ∠ABC + 2 ∠BCD (axiom)
∠AOC + ∠BOD = 2 × (∠ABC + ∠BCD)
∠AOC + ∠BOD = 2 × 90°
∠AOC + ∠BOD = 180°
∠AOC + ∠BOD = (1/2) × 360°
Hence, Arc APC + Arc BQD = (1/2) × Circumference (∠ at center is half of 360°)