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Geometry · Chapter 29

Angle Chasing

Straight lines, triangles, parallels, polygons and circles: five simple angle rules that, chained together, crack surprisingly hard contest problems.

You'll need

Nothing beyond solid middle-school math. This is where it all starts.

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29.1a0%

Lines, Triangles and Isosceles Chains

In this section you will be able to chase angles with straight lines, crossing lines, parallels, triangle sums and isosceles triangles.

A detective never sees the crime; she follows clues. Five rules (straight lines, triangles, parallels, polygons and circles) are all the clues you get.

The book’s three ground rules come first. Angles on a straight line add to 180∘180^\circ. Angles in a triangle add to 180∘180^\circ. And in a triangle, equal sides face equal angles (and equal angles face equal sides): that is an isosceles triangle.

Angle chasing means naming one unknown angle, say xx, and writing every other angle in terms of it. Look: an exterior angle of a triangle equals the sum of the two far angles. That one fact powers most chains.

Exploreplay first, formulas later

Triangle ABC: 73° + 64° + 43° = 180°

corners lined up: a straight line, 180°
73° + 64° + 43° = 180°

Tip: drag a colored point, or press Tab to reach it and use the arrow keys.

Angles 73°, 64° and 43°. Sum: 180°.

Name it

Concept 29.1.1 · Complementary Angles

Complementary angles are a pair of angles with a sum of 90∘90^\circ.

Concept 29.1.2 · Supplementary Angles

Supplementary angles are a pair of angles with a sum of 180∘180^\circ.

Concept 29.1.3 · Intersecting Lines

When two lines intersect, the vertical angles are equal. Vertical angles are each of the pairs of opposite angles made by two intersecting lines. "Vertical" here means they share the same vertex (corner point), not the usual meaning of up-down.

Concept 29.1.4 · Parallel Lines

When a line crosses two parallel lines, corresponding angles are equal. So every angle at the crossings is either aa or bb, with a+b=180∘a + b = 180^\circ.

Watch itpredict, then reveal

Example 29.1

AMC 10

Triangle ABCABC has AB=2⋅ACAB = 2 \cdot AC. Let DD and EE be on AB‾\overline{AB} and BC‾\overline{BC}, respectively, such that ∠BAE=∠ACD\angle BAE = \angle ACD. Let FF be the intersection of segments AEAE and CDCD, and suppose that △CFE\triangle CFE is equilateral. What is ∠ACB\angle ACB?

Geometry figure (its total is revealed at the end of the example)

Figure: triangle ABC; segment AE; segment CD; triangle CFE; angle EAB = x; angle DCA = x.

  1. 1

    Call the two equal angles xx: ∠BAE=∠ACD=x\angle BAE = \angle ACD = x. Triangle CFECFE is equilateral. What is ∠FCE\angle FCE?

    Predict first:
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Example 29.2

AIME

In △ABC\triangle ABC with AB=ACAB = AC, point DD lies strictly between AA and CC on side AC‾\overline{AC}, and point EE lies strictly between AA and BB on side AB‾\overline{AB} such that AE=ED=DB=BCAE = ED = DB = BC. The degree measure of ∠ABC\angle ABC is mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm + n.

Angle lab (its total is revealed at the end of the example)

links4
apex x22°
Angles in the chainx = 22°, 2x = 44°, 3x = 66°, 4x = 88°Not closed: AC ≠ AB.

Tip: drag a colored point, or press Tab to reach it and use the arrow keys.

Apex x = 22°. 4 equal links; their angles are x, 2x, 3x and so on. The chain does not close yet.

  1. 1

    Call the apex angle ∠A=x\angle A = x. Triangle AEDAED has AE=EDAE = ED. Which angle equals xx, and what is ∠ADE\angle ADE?

    Predict first:
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Try it

Problem 29.2.15(additional)

AMC 10

The sum of two angles of a triangle is 65\frac{6}{5} of a right angle, and one of these two angles is 30∘30^\circ larger than the other. What is the degree measure of the largest angle in the triangle?

Problem 29.2.10(additional)

AMC 10

This problem also appears in the book as 15.1.9.

How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?

Contest cornerreal contest problems

Problem 29.2.13(additional)

AHSME

In the adjoining figure, ABCDABCD is a square, ABEABE is an equilateral triangle and point EE is outside square ABCDABCD. What is the measure of ∡AED\measuredangle AED in degrees?

Problem 29.2.12(additional)

AMC 10/12

As shown in the figure below, point EE lies on the opposite half-plane determined by line CDCD from point AA so that ∠CDE=110∘\angle CDE = 110^\circ. Point FF lies on AD‾\overline{AD} so that DE=DFDE = DF, and ABCDABCD is a square. What is the degree measure of ∠AFE\angle AFE?

Problem 29.2.8(additional)

AHSME

△ABC\triangle ABC is isosceles with base ACAC. Points PP and QQ are respectively in CBCB and ABAB and such that AC=AP=PQ=QBAC = AP = PQ = QB. The number of degrees in ∠B\angle B is:

Problem 29.2.19(additional)

AMC 10Challenge

Quadrilateral ABCDABCD has AB=BC=CDAB = BC = CD, angle ABC=70∘ABC = 70^\circ and angle BCD=170∘BCD = 170^\circ. What is the measure of angle BADBAD?

Checkpoint for section 29.1a

Get 2 of 3 right to master this section. One try each, no hints.
1

△ABC\triangle ABC is isosceles with base ACAC. Points PP and QQ are respectively in CBCB and ABAB and such that AC=AP=PQ=QBAC = AP = PQ = QB. The number of degrees in ∠B\angle B is:

2

Two angles are supplementary, and one is 44 times the other. What is the smaller angle, in degrees?

3

In triangle ABCABC, AB=ACAB = AC and ∠B=50∘\angle B = 50^\circ. What is ∠A\angle A, in degrees?

29.1b0%

Polygon Angles

In this section you will be able to find interior and exterior angles of any polygon, and spot the ones worth memorising.

Walk once around any polygon and you turn exactly one full circle. That single fact gives you every exterior angle, and every interior angle with it.

Draw all the diagonals from one corner of an nn-gon. You get n−2n - 2 triangles, so the interior angles add to (n−2)⋅180∘(n - 2) \cdot 180^\circ. In a regular polygon all nn corners share that sum equally.

Look: the exterior angle is often the faster route. It is just 360∘÷n360^\circ \div n, and interior =180∘−= 180^\circ - exterior.

Exploreplay first, formulas later

Regular hexagon: interior angles add to (6 − 2) × 180° = 720°, each 120°, each exterior angle 360° ÷ 6 = 60°

sides6
4 triangles × 180° = 720°
Sum = 720°

Each corner: 720° ÷ 6 = 120°
Each exterior angle: 360° ÷ 6 = 60°

Walk around the edge. At each corner you turn by the exterior angle.

Regular hexagon: interior angles add to 720°, each 120°. Exterior angle 60°.

Name it

Theorem 29.1.5

For a polygon with nn sides:

Sum of interior angles=(n−2)⋅180∘\text{Sum of interior angles} = (n-2)\cdot 180^\circ
Interior angle of a regular polygon=(n−2)n⋅180∘\text{Interior angle of a regular polygon} = \frac{(n-2)}{n}\cdot 180^\circ
Exterior angle of a regular polygon=360∘n\text{Exterior angle of a regular polygon} = \frac{360^\circ}{n}

Fact 29.1.6 · Important Interior Angles

Important interior angles of regular polygons:

Number of sides Interior angle
3 60∘60^\circ
4 90∘90^\circ
5 108∘108^\circ
6 120∘120^\circ
8 135∘135^\circ
9 140∘140^\circ
10 144∘144^\circ

Try it

Problem 29.2.14(additional)

AMC 10

The angles of quadrilateral ABCDABCD satisfy ∠A=2∠B=3∠C=4∠D\angle A = 2\angle B = 3\angle C = 4\angle D. What is the degree measure of ∠A\angle A, rounded to the nearest whole number?

Problem 29.2.16(additional)

AMC 10

This problem also appears in the book as 15.1.8.

Mary divides a circle into 1212 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

Contest cornerreal contest problems

Problem 29.2.4

AMC 10

Square ABCDABCD is rotated 20∘20^\circ clockwise about its center to obtain square EFGHEFGH, as shown below. What is the degree measure of ∠EAB\angle EAB?

Problem 29.2.5

MATHCOUNTS

A square is located in the interior of a regular hexagon, and certain vertices are labeled as shown. What is the degree measure of ∠ABC\angle ABC?

Problem 29.2.7

AMC 10

The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with 99 trapezoids, let xx be the angle measure in degrees of the larger interior angle of the trapezoid. What is xx?

Checkpoint for section 29.1b

Get 2 of 3 right to master this section. One try each, no hints.
1

What is the sum of the interior angles of a heptagon (7 sides), in degrees?

2

Each exterior angle of a regular polygon is 24∘24^\circ. How many sides does it have?

3

Each interior angle of a regular polygon is 150∘150^\circ. How many sides does it have?

29.1c0%

Angles in Circles

In this section you will be able to use inscribed angles, chords, tangents and cyclic quadrilaterals to chase angles in circles.

Stand anywhere on the far side of a circle and look at a fixed arc. You see it at the same angle from every spot: exactly half of the angle at the center.

Circles hand you equal angles for free. Every angle in this section comes from one idea: an angle is half the arc it looks at. Chords crossing inside average two arcs; secants meeting outside take half their difference; a tangent and a chord take half the arc between them.

Look: radii are all equal, so drawing them creates isosceles triangles. And a radius meets a tangent at 90∘90^\circ.

Exploreplay first, formulas later

Inscribed angle APB = 55°, half of arc AB = 110°

∠APB = 55°

Arc AB (the part not touching P) = 110°. The angle at P is half of it, wherever P sits on the other side.

Inscribed angle APB = 55°. Arc AB = 110°, which is twice as big.

Name it

Theorem 29.1.7 · Inscribed Arc Theorem

The angle an arc makes at the center (the arc's angle) is double the angle it makes at any point on the edge of the circle.

central angle=2×inscribed angle\text{central angle} = 2 \times \text{inscribed angle}

Corollary 29.1.8 · Inscribed Right Triangle

An inscribed triangle with a diameter as one side is always a right triangle.

Definition 29.1.9 · Chord

A chord is a line segment between any two distinct points on the circle. The diameter of the circle is the longest chord in the circle.

Theorem 29.1.10

The perpendicular bisector of any chord passes through the center. In the figure, the perpendicular bisectors of ABAB and CDCD intersect at the center OO.

The perpendicular bisectors of chords AB and CD both pass through the center O

Both dashed bisectors meet at O

Drag A, B, C or D. However you tilt the chords, the dashed lines cross exactly at the center.

The perpendicular bisectors of AB and CD both pass through the center O.

Corollary 29.1.11

  • Congruent chords are equidistant from the center of a circle.
  • If two chords in a circle are congruent, then their intercepted arcs are congruent.
  • If two chords in a circle are congruent, then they determine two central angles that are congruent.

Equal chords AB and CD cut off equal arcs of 65° and sit the same distance from O

∠AOB = ∠COD = 65°

Drag C to slide the copy of chord AB around. Its distance from O stays 84.3, the same as AB's.

AB = CD. Both central angles are 65°, both arcs are 65°, and both chords are 84.3 units from O.

Theorem 29.1.12

Two secants from an outside point PP cut the circle at A,BA, B and C,DC, D. The angle at PP is half the difference of the two arcs it cuts off:

∠APC=BD⌢−AC⌢2\angle APC = \frac{\overset{\frown}{BD} - \overset{\frown}{AC}}{2}

Two secants from P: angle APC = (arc BD − arc AC) ÷ 2 = (150° − 40°) ÷ 2 = 55°

∠APC = (150° − 40°) ÷ 2 = 55°

The far arc BD pushes the angle open; the near arc AC closes it.

Angle APC = 55°. Arc BD = 150°, arc AC = 40°, half the difference = 55°.

Theorem 29.1.13

If two chords ABAB and CDCD intersect at PP, then ∠BPC\angle BPC and ∠APD\angle APD are equal to the average of the two arcs:

∠BPC=∠APD=BC⌢+AD⌢2\angle BPC = \angle APD = \frac{\overset{\frown}{BC} + \overset{\frown}{AD}}{2}

Chords AB and CD cross at P: angle BPC = (arc BC + arc AD) ÷ 2 = (135° + 75°) ÷ 2 = 105°

∠BPC = (135° + 75°) ÷ 2 = 105°

Each angle at P looks at two arcs: the one in front and the one behind. Average them.

Angle BPC = 105°. Arc BC = 135°, arc AD = 75°, their average = 105°.

Theorem 29.1.14

If a tangent line through RR touches the circle at QQ, and a chord QPQP is drawn, then ∠RQP\angle RQP is equal to half the arc QPQP inside the angle.

Tangent at Q and chord QP: angle RQP = arc QP ÷ 2 = 170° ÷ 2 = 85°

∠RQP = 170° ÷ 2 = 85°

The red arc is the part of the circle inside the angle. Slide P and watch both numbers.

Angle RQP = 85°, half of arc QP = 170°.

Remark 29.1.15

Circles are really useful for angle chasing, so keep an eye out for the inscribed arc theorem. It can be used in many angle chasing problems.

Remark 29.1.16

A useful trick for angle chasing problems with regular polygons: draw a circle around the polygon and use the inscribed arc theorem.

Inscribed angle AEC = 60°, half of arc AC = 120°

∠AEC = 60°

Arc AC (the part not touching E) = 120°. The angle at E is half of it, wherever E sits on the other side.

Inscribed angle AEC = 60°. Arc AC = 120°, which is twice as big.

Theorem 29.1.17

Equal chords mark out equal arcs. This means that if you have 2 chords of the same length, the arcs (and sectors) of the circle they mark out are equal.

Definition 29.1.18 · Tangent

A tangent is any line from a point external to the circle that just touches the circle.

Theorem 29.1.19 · Right Angle Tangency Point

If you connect the center of a circle to the point where the circle and a line are tangent, they form a right angle.

Tangents PS and PT: the radius meets each tangent at 90°, and PS = PT

∠OSP = ∠OTP = 90°

Drag P anywhere outside. The radius always meets the tangent at a right angle, and both tangents stay 118.4 long.

Angle OSP = 90° and angle OTP = 90°. PS = PT = 118.4.

Theorem 29.1.21 · Properties of a Cyclic Quadrilateral

In a cyclic quadrilateral (all four corners on one circle), the sum of opposite angles is 180∘180^\circ: p+q=180∘p + q = 180^\circ and r+s=180∘r + s = 180^\circ.

Watch itpredict, then reveal

Example 29.3

AMC 10

Let △ABC\triangle ABC be an isosceles triangle with BC=ACBC = AC and ∠ACB=40∘\angle ACB = 40^\circ. Construct the circle with diameter BC‾\overline{BC}, and let DD and EE be the other intersection points of the circle with the sides AC‾\overline{AC} and AB‾\overline{AB}, respectively. Let FF be the intersection of the diagonals of the quadrilateral BCDEBCDE. What is the degree measure of ∠BFC\angle BFC?

Inscribed angle (its total is revealed at the end of the example)

∠BDC = ?

Arc BC (the part not touching D) = ?. Can you predict the angle before you look?

Inscribed angle BDC: predict it first.

  1. 1

    BCBC is a diameter and DD is on the circle. What is ∠BDC\angle BDC? (And ∠BEC\angle BEC?)

    Predict first:
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Contest cornerreal contest problems

Problem 29.2.2

AMC

Two congruent circles centered at points AA and BB each pass through the other circle's center. The line containing both AA and BB is extended to intersect the circles at points CC and DD. The circles intersect at two points, one of which is EE. What is the degree measure of ∠CED\angle CED?

Problem 29.2.6

AMC 10

In the given circle, the diameter EB‾\overline{EB} is parallel to DC‾\overline{DC}, and AB‾\overline{AB} is parallel to ED‾\overline{ED}. The angles AEBAEB and ABEABE are in the ratio 4:54 : 5. What is the degree measure of angle BCDBCD?

Problem 29.2.3

AMC 10

Abdul and Chiang are standing 48 feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chiang measures 60∘60^\circ. What is the square of the distance (in feet) between Abdul and Bharat?

Problem 29.2.17(additional)

AHSME

An acute isosceles triangle, ABCABC, is inscribed in a circle. Through BB and CC, tangents to the circle are drawn, meeting at point DD. If ∠ABC=∠ACB=2∠D\angle ABC = \angle ACB = 2\angle D and xx is the radian measure of ∠A\angle A, then x=x =

Checkpoint for section 29.1c

Get 2 of 3 right to master this section. One try each, no hints.
1

Points AA, BB, CC are on a circle with center OO, and ∠AOB=110∘\angle AOB = 110^\circ. CC is on the major arc ABAB. What is ∠ACB\angle ACB, in degrees?

2

ABCDABCD is a quadrilateral inscribed in a circle, and ∠A=72∘\angle A = 72^\circ. What is ∠C\angle C, in degrees?

3

Chords ABAB and CDCD of a circle cross at PP. The arcs cut off by ∠APD\angle APD and its vertical angle are AD⌢=50∘\overset{\frown}{AD} = 50^\circ and BC⌢=90∘\overset{\frown}{BC} = 90^\circ. What is ∠APD\angle APD, in degrees?

29.2

Practice Problems

In this section you will be able to mix every angle-chasing tool from this chapter on harder contest problems.

These problems need several clues at once: polygons, isosceles chains, symmetry and even light beams. Every problem has three hints if you get stuck.

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 29.2.1

AMC 10

Let ABCDEFABCDEF be an equiangular hexagon. The lines AB,CD,AB, CD, and EFEF determine a triangle with area 1923192\sqrt{3}, and the lines BC,DE,BC, DE, and FAFA determine a triangle with area 3243324\sqrt{3}. The perimeter of hexagon ABCDEFABCDEF can be expressed as m+npm + n\sqrt{p}, where m,n,m, n, and pp are positive integers and pp is not divisible by the square of any prime. What is m+n+pm + n + p?

Equiangular hexagon ABCDEF (every angle 120°) with sides 14.76, 6.48, 14.76, 6.48, 14.76, 6.48; perimeter ?; lines AB, CD, EF make an equilateral triangle of side 27.71; lines BC, DE, FA make an equilateral triangle of side 36

ABCDEF
Perimeter = ? (hidden until the end of the example)
Triangle (AB, CD, EF): side 27.71, area 332.55
Triangle (BC, DE, FA): side 36, area 561.18

Equiangular hexagon ABCDEF (every angle 120°) with sides 14.76, 6.48, 14.76, 6.48, 14.76, 6.48; perimeter ?; lines AB, CD, EF make an equilateral triangle of side 27.71; lines BC, DE, FA make an equilateral triangle of side 36

Problem 29.2.9(additional)

AHSME

In this diagram ABAB and ACAC are the equal sides of an isosceles △ABC\triangle ABC, in which is inscribed equilateral △DEF\triangle DEF. Designate ∠BFD\angle BFD by aa, ∠ADE\angle ADE by bb, and ∠FEC\angle FEC by cc. Then:

Click a choice or type its letter (A–E).

Problem 29.2.11(additional)

MATHCOUNTS

Concave quadrilateral ABCDABCD is symmetric about the line ACAC. The measures of angles DABDAB and ABCABC are 8484 degrees and 3232 degrees, respectively. The dashed line segments bisect angles ABCABC and ADCADC. What is the degree measure of the acute angle at which the two dashed line segments intersect?

Problem 29.2.18(additional)

AIMEChallenge

Triangle ABCABC is isosceles with AC=BCAC = BC and ∠ACB=106∘\angle ACB = 106^\circ. Point MM is in the interior of the triangle so that ∠MAC=7∘\angle MAC = 7^\circ and ∠MCA=23∘\angle MCA = 23^\circ. Find the number of degrees in ∠CMB\angle CMB.

Problem 29.2.20(additional)

AIMEChallenge

A beam of light strikes BC‾\overline{BC} at point CC with angle of incidence α=19.94∘\alpha = 19.94^\circ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments AB‾\overline{AB} and BC‾\overline{BC} according to the rule: angle of incidence equals angle of reflection. Given that β=α/10=1.994∘\beta = \alpha/10 = 1.994^\circ and AB=BCAB = BC, determine the number of times the light beam will bounce off the two line segments. Include the first reflection at CC in your count.