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
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
