Given: quadrilateral abcd is inscribed in circle o.
prove: m∠a + m∠c = 180°

drag an expression or phrase to each box to complete the proof.

statements → reasons
1. → given
2. mbcd = 2(m∠a) →
3. mdab = 2(m∠c) → inscribed angle theorem
4. → the sum of arcs that make a circle is 360°.
5. 2(m∠a) + 2(m∠c) = 360° →
6. m∠a + m∠c = 180° → division property of equality

answer choices:
substitution property
inscribed angle theorem
central angle theorem
mbcd + mdab = 360°
mbcd = mdab
quadrilateral abcd is inscribed in circle o.

i'm guessing:
1. quadrilateral abcd is inscribed in circle o.
4. mbcd + mdab = 360°
5. inscribed angle theorem

i'm not sure about 5 or 2.

.

1. Quadrilateral ABCD is inscribed in circle O
A quadrilateral is a four sided figure, in this case ABCD is a cyclic quadrilateral such that all its vertices touches the circumference of the circle.
A cyclic quadrilateral is a four sided figure with all its vertices touching the circumference of a circle.

2. mBCD = 2 (m∠A) = Inscribed Angle Theorem
An inscribed angle is an angle with its vertex on the circle, formed by two intersecting chords. 
Such that Inscribed angle = 1/2 Intercepted Arc
In this case the inscribed angle is m∠A and the intercepted arc is MBCD
Therefore; m∠A = 1/2 mBCD

4. The sum of arcs that make up a circle is 360
Therefore; mBCD + mDAB = 360°
The circles is made up of arc BCD and arc DAB, therefore the sum angle of the arcs is equivalent to 360°

5. 2(m∠A + 2(m∠C) = 360; this is substitution property
From step 4 we stated that mBCD +mDAB = 360
but from the inscribed angle theorem;
mBCD= 2 (m∠A) and mDAB = 2(m∠C)
Therefore; substituting in the equation in step 4 we get;
2(m∠A) + 2(m∠C) = 360