Question

Given: quadrilateral ABCD inscribed in a circle Prove: ∠A and ∠C are supplementary, ∠B and ∠D are supplementary Let the measure of = a°. Because and form a circle, and a circle measures 360°, the measure of is 360 – a°. Because of the ________ theorem, m∠A = degrees and m∠C = degrees. The sum of the measures of angles A and C is degrees, which is equal to , or 180°. Therefore, angles A and C are supplementary because their measures add up to 180°. Angles B and D are supplementary because the sum of the measures of the angles in a quadrilateral is 360°. m∠A + m∠C + m∠B + m∠D = 360°, and using substitution, 180° + m∠B + m∠D = 360°, so m∠B + m∠D = 180°. What is the missing information in the paragraph proof?

Answer 1

Quadrilateral ABCD is inscribed in a circle. Let the measure of arc BAD be a°. Arcs BCD and BAD form a circle  and a circle measures 360°, then  measure of arc BCD is 360°-a°.

The inscribed angle theorem states that an angle  inscribed in a circle is half of the central angle  that subtends the same arc on the circle.

Because of the Inscribed angle theorem,

$m\angle A=\dfrac{a^{\circ}}{2};$

$m\angle C=\dfrac{360^{\circ}-a^{\circ}}{2}.$

The sum of the measures of angles A and C is

$\dfrac{a^{\circ}}{2}+\dfrac{360^{\circ}-a^{\circ}}{2}=180^{\circ}.$

Therefore, angles A and C are supplementary, because their measures add up to 180°.

Angles B and D are supplementary, because the sum of the measures of the angles in a quadrilateral is 360°.

m∠A + m∠C + m∠B + m∠D = 360°,

and using substitution,

180° + m∠B + m∠D = 360°, so m∠B + m∠D = 180°.

Answer: inscribed angle theorem

Answer 2

Solution:

Given: quadrilateral ABCD inscribed in a circle

To Prove:

1. ∠A and ∠C are supplementary.

2. ∠B and ∠D are supplementary.

Construction : Join AC and BD.

Proof: As, angle in same segment of circle are equal.Considering AB, BC, CD and DA as Segments, which are inside the circle,

∠1=∠2-----(1)

∠3=∠4-----(2)

∠5=∠6-------(3)

∠7=∠8------(4)

Also, sum of angles of quadrilateral is 360°.

⇒∠A+∠B+∠C+∠D=360°

→→∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360°→→→using 1,2,3,and 4

→→→2∠1+2∠4+2∠6+2∠8=360°

→→→→2( ∠1 +∠6) +2(∠4+∠8)=360°⇒Dividing both sides by 2,

→→→∠B + ∠D=180°as, ∠1 +∠6=∠B , ∠4+∠8=∠B------(A)

As, ∠A+∠B+∠C+∠D=360°

∠A+∠C+180°=360°

∠A+∠C=360°-180°------Using A

∠A+∠C=180°

Hence proved.