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Cyclic Quadrilaterals: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 An opposite whole angle

Convex cyclic quadrilateral $ABCD$ has $m\angle ABC=118^\circ$ . Find the whole angle $CDA$ and justify the relationship you use.

Schematic; use the stated givens.

Text description of this figure

A convex cyclic quadrilateral has angle B given and the opposite whole angle at D unknown.
1. Hint 1
  Identify the vertex opposite B.
2. Hint 2
  The two whole opposite angles must total a straight angle.
Answer

$m\angle CDA=62^\circ$ .

Full solution

The opposite vertices are B and D.
Since all four vertices lie on one circle in convex order, the opposite-angle theorem applies.

Subtract the given angle from a straight angle.
$\begin{aligned}m\angle CDA&=180^\circ-118^\circ,\\m\angle CDA&=62^\circ.\end{aligned}$

Answer

$m\angle CDA=62^\circ$ .

Key idea

Use the whole angle at the opposite vertex.
Problem 2 Determine the angle expressions

In convex cyclic quadrilateral $PQRS$ , whole angles at P,R measure $(3x+2)^\circ$ and $(6x-11)^\circ$ . Find x and both whole angle measures.

Schematic; use the stated givens.

Text description of this figure

P and R are opposite vertices of a convex cyclic quadrilateral.
1. Hint 1
  P and R are opposite vertices.
2. Hint 2
  Add the two expressions and set their sum equal to 180 degrees.
3. Hint 3
  Substitute your value into each original expression and check that both angles are positive.
Answer

$\begin{gathered}x=21,\\m\angle P=65^\circ,\\m\angle R=115^\circ.\end{gathered}$

Full solution

Opposite angles are supplementary.
$\begin{aligned}(3x+2)+(6x-11)&=180,\\9x-9&=180,\\x&=21.\end{aligned}$

Substitute into the angle at P.
$\begin{aligned}m\angle P&=(3(21)+2)^\circ,\\m\angle P&=65^\circ.\end{aligned}$

Substitute into the angle at R.
$\begin{aligned}m\angle R&=(6(21)-11)^\circ,\\m\angle R&=115^\circ.\end{aligned}$

Both angles lie strictly between zero and 180 degrees, and 65 plus 115 is 180.

Answer

$\begin{gathered}x=21,\\m\angle P=65^\circ,\\m\angle R=115^\circ.\end{gathered}$

Key idea

An algebraic answer must give valid angle measures as well as the required sum.
Problem 3 An exterior angle and two pairs

Convex cyclic $ABCD$ has side BC extended through C to E. Given $m\angle DCE=107^\circ$ and $m\angle CDA=86^\circ$ , find the whole interior angles at A,B,C.

Schematic; use the stated givens.

Text description of this figure

Convex cyclic ABCD has B,C,E collinear in that order; the exterior angle at C is DCE.
1. Hint 1
  The exterior angle at C is paired with the opposite interior angle.
2. Hint 2
  At C, the interior and exterior angles form a linear pair.
3. Hint 3
  Use the opposite pair B,D to find the remaining requested angle.
Answer

$\begin{gathered}m\angle A=107^\circ,\\m\angle B=94^\circ,\\m\angle C=73^\circ.\end{gathered}$

Full solution

The exterior angle equals the opposite interior angle, so the whole angle at A is 107 degrees.

The linear pair at C gives
$m\angle C=180^\circ-107^\circ.$
Thus the whole angle at C is 73 degrees.

The opposite pair B,D gives
$m\angle B=180^\circ-86^\circ.$
Thus the whole angle at B is 94 degrees.

Answer

$\begin{gathered}m\angle A=107^\circ,\\m\angle B=94^\circ,\\m\angle C=73^\circ.\end{gathered}$

Key idea

An exterior angle is equal to the opposite interior angle and supplementary to its neighboring interior angle.
Problem 4 Two cyclicity decisions

Each case describes a convex quadrilateral ABCD with four distinct vertices in boundary order and no three collinear. Decide whether it is cyclic and justify your answer.

Case A: $m\angle DAB=88^\circ$ and $m\angle BCD=92^\circ$ .

Case B: $m\angle ABC=64^\circ$ and $m\angle CDA=114^\circ$ .
1. Hint 1
  For each case, identify whether the given pair is opposite.
2. Hint 2
  A sum of exactly 180 degrees proves cyclicity; a different opposite sum rules it out.
Answer

Case A is cyclic. Case B is not cyclic.

Full solution

Case A: the opposite angles at A,C total 180 degrees.
The converse cyclicity test proves that all four vertices share a circle.

Case B: the opposite angles at B,D total 178 degrees.
The opposite-angle theorem would require 180 degrees if the quadrilateral were cyclic, so it is not cyclic.

Answer

Case A is cyclic. Case B is not cyclic.

Key idea

Distinguish using a converse to establish a circle from using a necessary condition to rule one out.
Problem 5 Separate the diagonal pieces

Convex cyclic quadrilateral ABCD has diagonal BD. Given $m\angle BAD=99^\circ$ , $m\angle ABD=28^\circ$ , and $m\angle DBC=37^\circ$ , find the whole angle BCD and the angle part BDC.

Schematic; use the stated givens.

Text description of this figure

Diagonal BD splits the whole angles at B and D in convex cyclic ABCD.
1. Hint 1
  Find the whole angle at C from its opposite angle A.
2. Hint 2
  Triangle ABD gives one part of the whole angle at D.
3. Hint 3
  Add the two pieces at B, find its opposite whole angle at D, and subtract the known part there.
Answer

$\begin{gathered}m\angle BCD=81^\circ,\\m\angle BDC=62^\circ.\end{gathered}$

Full solution

The whole angle at C is opposite A.
$m\angle BCD=180^\circ-99^\circ.$
It measures 81 degrees.

Triangle ABD gives the first part at D.
$\begin{aligned}m\angle ADB&=180^\circ-99^\circ-28^\circ,\\m\angle ADB&=53^\circ.\end{aligned}$

The whole angle at B is the sum of its two given parts, or 65 degrees.
Its opposite whole angle CDA is therefore 115 degrees.

Diagonal BD lies inside the whole angle at D.
Subtract its part ADB.
$\begin{aligned}m\angle BDC&=115^\circ-53^\circ,\\m\angle BDC&=62^\circ.\end{aligned}$

Triangle BCD checks the result: its angles 37, 81, and 62 degrees total 180 degrees.

Answer

$\begin{gathered}m\angle BCD=81^\circ,\\m\angle BDC=62^\circ.\end{gathered}$

Key idea

Keep whole opposite angles separate from the parts made by a diagonal.
Problem 6 A consecutive pair is not enough

Does having one supplementary pair of consecutive angles force a convex quadrilateral to be cyclic? Refute the statement with one exact convex example. Give a construction or angle data that can actually occur, and justify both the supplementary consecutive pair and the failure of cyclicity.
1. Hint 1
  Consider a parallelogram with no right angles.
2. Hint 2
  Choose a nonright angle and its supplement. A parallelogram supplies equal opposite angles.
3. Hint 3
  Test an opposite pair in your example rather than only the given consecutive pair.
Answer

No. One example is a parallelogram with boundary-order angles $72^\circ,108^\circ,72^\circ,108^\circ$ .

Full solution

Draw segment AB of length 6 units.
At A construct ray AD making a 72-degree angle with ray AB, and put D 4 units from A on it.
Through B draw a parallel to AD and through D a parallel to AB; their intersection is C.
This gives a nondegenerate convex parallelogram ABCD.

Its angle at A is 72 degrees.
Parallelogram angle properties give successive angles of 72, 108, 72, and 108 degrees.
Thus consecutive angles A,B total 180 degrees.

Opposite angles A,C total 144 degrees.
This violates the necessary opposite-angle condition, so the example is not cyclic.
A single valid example refutes the proposed implication.

Schematic; use the stated givens.

Answer

No. One example is a parallelogram with boundary-order angles $72^\circ,108^\circ,72^\circ,108^\circ$ .

Key idea

A converse needs the exact hypothesis of the theorem, including which vertices are opposite.
Problem 7 An exterior-angle cyclicity test

Let ABCD be a convex quadrilateral with four distinct vertices in boundary order and no three collinear. Extend BC through C to E. Prove both directions of this statement: ABCD is cyclic if and only if $m\angle DCE=m\angle BAD$ .

Schematic; use the stated givens.

Text description of this figure

No circle is assumed. Side BC is extended through C to E in a convex quadrilateral.
1. Hint 1
  Start with the linear pair at C, which holds whether or not there is a circle.
2. Hint 2
  In the forward direction, compare the two supplements of angle BCD.
3. Hint 3
  In the reverse direction, replace the exterior angle in the linear-pair sum and then apply the opposite-angle converse.
Answer

Both directions hold under the stated convexity and extension conditions.

Full solution

Forward direction: suppose ABCD is cyclic.
The exterior angle DCE is supplementary to angle BCD because B,C,E are in order on one line.
The opposite-angle theorem makes BAD supplementary to BCD as well.
Therefore DCE and BAD are equal.

Reverse direction: suppose DCE and BAD are equal.
The linear pair gives
$m\angle DCE+m\angle BCD=180^\circ.$
Replace DCE by equal angle BAD to obtain a supplementary opposite pair.

The converse opposite-angle test applies to this convex quadrilateral, proving that ABCD is cyclic.

Answer

Both directions hold under the stated convexity and extension conditions.

Key idea

The same linear pair connects the exterior equality to the interior opposite-angle test.
Problem 8 Equal opposite angles

Convex quadrilateral ABCD has equal whole angles at A and C. Its other whole angles satisfy $m\angle ABC=104^\circ$ and $m\angle CDA=76^\circ$ . Determine whether ABCD is cyclic, find the whole angles at A,C, and identify a diameter of its circle if cyclic.
1. Hint 1
  The known angles form an opposite pair.
2. Hint 2
  Use the quadrilateral angle sum and the equality of the two unknown whole angles.
3. Hint 3
  A right inscribed angle intercepts a semicircle, whose endpoints determine the diameter opposite that vertex.
Answer

ABCD is cyclic, and BD is a diameter.
$\begin{gathered}m\angle A=90^\circ,\\m\angle C=90^\circ.\end{gathered}$

Full solution

The opposite angles at B,D total 180 degrees, so the converse makes ABCD cyclic.

The quadrilateral sum leaves 180 degrees for the equal angles A,C.
Each therefore measures 90 degrees.

The right inscribed angle at A has endpoints B,D.
Its intercepted arc is a semicircle, so BD is a diameter by the semicircle converse.

Schematic; use the stated givens.

Answer

ABCD is cyclic, and BD is a diameter.
$\begin{gathered}m\angle A=90^\circ,\\m\angle C=90^\circ.\end{gathered}$

Key idea

Equal opposite angles become right angles once their supplementary sum is known.
Problem 9 Transfer from parallel bases

Convex isosceles trapezoid ABCD has chosen parallel bases AB and CD. The whole angle DAB is 74 degrees. Find all four whole interior angles and decide whether the trapezoid is cyclic, giving a geometric justification.
1. Hint 1
  An isosceles trapezoid has equal angles at each chosen base.
2. Hint 2
  A leg cuts the parallel bases, so its same-side interior angles are supplementary.
3. Hint 3
  After finding the angles, test a pair of opposite vertices.
Answer

The trapezoid is cyclic.
$\begin{gathered}m\angle A=74^\circ,\\m\angle B=74^\circ,\\m\angle C=106^\circ,\\m\angle D=106^\circ.\end{gathered}$

Full solution

The equal base angles at AB give angle B equal to angle A, or 74 degrees.

The parallel bases make B,C supplementary along leg BC.
Thus C is 106 degrees.
The equal base angles at CD give D the same measure.

Opposite angles A,C total 180 degrees.
The vertices are distinct and the trapezoid is convex, so the converse proves cyclicity.

Answer

The trapezoid is cyclic.
$\begin{gathered}m\angle A=74^\circ,\\m\angle B=74^\circ,\\m\angle C=106^\circ,\\m\angle D=106^\circ.\end{gathered}$

Key idea

Earlier parallel-line and quadrilateral facts can supply a circle test without a circle in the givens.
Problem 10 Decide first, then transfer

Convex quadrilateral ABCD has both diagonals AC and BD. Triangle ABC has $m\angle BAC=28^\circ$ and $m\angle ABC=79^\circ$ . The other angle parts are $m\angle CAD=34^\circ$ and $m\angle ACD=45^\circ$ .

Determine whether ABCD is cyclic with a complete geometric justification. Then find $m\angle CBD$ and the whole angle ADC.

Schematic; use the stated givens.

Text description of this figure

The givens form a convex quadrilateral with both diagonals but no circle; all angle values are stated in the text.
1. Hint 1
  Find the missing angle of triangle ABC.
2. Hint 2
  Add the diagonal pieces at A and at C before testing their opposite sum.
3. Hint 3
  After the circle is justified, compare angles CAD and CBD and use the opposite pair B,D.
Answer

ABCD is cyclic.
$\begin{gathered}m\angle CBD=34^\circ,\\m\angle ADC=101^\circ.\end{gathered}$

Full solution

Triangle ABC gives its missing angle.
$\begin{aligned}m\angle BCA&=180^\circ-28^\circ-79^\circ,\\m\angle BCA&=73^\circ.\end{aligned}$

Convexity puts AC inside both whole angles at A,C.
Add the corresponding pieces.
$\begin{gathered}m\angle BAD=62^\circ,\\m\angle BCD=118^\circ.\end{gathered}$

These opposite angles total 180 degrees.
The quadrilateral is convex with distinct vertices and no collinear triple, so the converse proves that A,B,C,D lie on one circle.

Angles CAD and CBD both intercept the arc CD excluding A,B.
They have equal measure, so CBD is 34 degrees.

The whole angle at D is opposite the given whole angle at B.
$\begin{aligned}m\angle ADC&=180^\circ-79^\circ,\\m\angle ADC&=101^\circ.\end{aligned}$

Answer

ABCD is cyclic.
$\begin{gathered}m\angle CBD=34^\circ,\\m\angle ADC=101^\circ.\end{gathered}$

Key idea

Establish cyclicity from whole opposite angles before transferring a diagonal angle.

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