Segment BD – diameter of a circle with center O. Chord AC bisects the radius of OB and is perpendicular

Segment BD – diameter of a circle with center O. Chord AC bisects the radius of OB and is perpendicular to it. Find the angles of the quadrangle ABCD and the degree measures of the arcs AB, BC, CD, AD.

1. Since BD is a diameter, it divides the circle into two arcs BAD = 180 ° and BCD = 180 °. Inscribed angles A and C are based on arcs BAD and BCD. The degree measure of the inscribed angle is equal to half the degree measure of the arc on which it rests, then:
∠А = arc BAD / 2 = 180 ° / 2 = 90 °;
∠C = arc BCD / 2 = 180 ° / 2 = 90 °.
2. From the center of the circle draw the radius to point A. Consider △ ВAO: ОА = ОВ. Since the chord AC is perpendicular to BD, the segment AH in △ BAO is the height. Since AН divides the OB in half, AН is the median. Thus, AH is both the median and the height ⇒ in △ VAO AB = AO, and also AO = OB, then AAO is a regular triangle. All angles of a regular triangle are 60 °: ∠BAO = 60 °, ∠AOB = 60 °, ∠OBA = 60 °.
Then AН is the bisector ∠BAO ⇒ ∠ВAН = ∠ОАН = ∠BAO / 2 = 60 ° / 2 = 30 °.
∠ВAН = ∠BAС = 30 °.
The inscribed ∠ BAC rests on the BC arc, then the degree measure of the BC arc is 2 times greater than the degree measure ∠ BAC ⇒ BC arc = 2 * ∠BAC = 2 * 30 ° = 60 °.
3. In △ BCD, a similar situation: CH – and the height, and the median, and the bisector, ∠BCA = 30 ° ⇒ arc AB = 2 * ∠BCA = 2 * 30 ° = 60 °.
4. Consider △ ABC: ∠BAC = ∠BCA = 30 ° ⇒ △ ABC – isosceles, ∠BAC and ∠BCA – angles at the base. By the theorem on the sum of the angles of a triangle:
∠BAC + ∠BCA + ∠ABS = 180 °;
30 ° + 30 ° + ∠ABS = 180 °;
∠ABS = 180 ° – 60 °;
∠ABS = 120 °.
∠ABS = ∠В = 120 °.
5. By the sum theorem, the catch of a quadrangle:
∠А + ∠В + ∠С + ∠D = 360 °;
90 ° + 120 ° + 90 ° + ∠D = 360 °;
∠D = 360 ° – 300 °;
∠D = 60 °.
6. ∠А consists of two corners ∠BAC and ∠CAD:
∠А = ∠ВАС + ∠CAD;
30 ° + ∠CAD = 90 °;
∠CAD = 90 ° – 30 °;
∠CAD = 60 °.
The inscribed ∠CAD rests on the CD arc ⇒ CD arc = 2 * ∠CAD = 2 * 60 ° = 120 °.
7.∠С consists of two angles ∠BCA and ∠AСD:
∠С = ∠ВСА + ∠АСD;
30 ° + ∠АСD = 90 °;
∠АСD = 90 ° – 30 °;
∠АСD = 60 °.
The inscribed ∠АСD rests on the arc АD ⇒ arc АD = 2 * ∠АСD = 2 * 60 ° = 120 °.
Answer: ∠А = 90 °, ∠В = 120 °, ∠С = 90 °, ∠D = 60 °. Arc AB = 60 °, BC = 60 °, CD = 120 °, AD = 120 °.