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 → Reasons1. ___________ → Given2. mBCD = 2(m∠A) → ________3. mDAB = 2(m∠C) → Inscribed Angle Theorem4. _________ → 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 EqualityAnswer Choices:Substitution PropertyInscribed Angle TheoremCentral Angle TheoremmBCD + mDAB = 360°mBCD = mDABQuadrilateral ABCD is inscribed in circle O.I'm guessing:1. Quadrilateral ABCD is inscribed in circle O.4. mBCD + mDAB = 360°5. Inscribed Angle TheoremI'm not sure about 5 or 2.Thanks.

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