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Circle Theorems: Question 2

Syllabus E4.7, E4.8

Structured Extended 7 marks

A, B, C and D are four points on a circle, labelled in order around the circumference. The chords AC and BD are drawn and intersect at the point E inside the circle.

You are given that

  • angle BAC = 47°
  • angle CAD = 61°
  • angle ABD = 31°

(a) Find angle BDC, giving a reason for your answer. [2]

(b) Write down the size of angle DBC, giving a reason for your answer. [1]

(c) Hence find angle ADC. [2]

(d) Find angle BCD. [2]

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Worked solution

Setting up

Because A, B, C, DA,\,B,\,C,\,D lie on the circle in that order, ABCDABCD is a cyclic quadrilateral whose diagonals are the chords ACAC and BDBD. Two circle theorems do all the work:

  • Same segment: angles standing on the same chord, from the same side, are equal.
  • Cyclic quadrilateral: opposite angles add up to 180∘180^\circ.

(a) angle BDCBDC

Chord BCBC subtends ∠BAC\angle BAC at AA and ∠BDC\angle BDC at DD. Since AA and DD lie on the same arc (the major arc BCBC), these are angles in the same segment:

∠BDC=∠BAC=47∘.\angle BDC = \angle BAC = 47^\circ.

Reason: angles in the same segment (standing on chord BCBC) are equal.

(b) angle DBCDBC

Chord DCDC subtends ∠DAC\angle DAC at AA and ∠DBC\angle DBC at BB, with AA and BB in the same segment:

∠DBC=∠DAC=61∘.\angle DBC = \angle DAC = 61^\circ.

Reason: angles in the same segment (standing on chord DCDC) are equal.

(c) angle ADCADC

First build the whole angle at BB using the result from (b):

∠ABC=∠ABD+∠DBC=31∘+61∘=92∘.\angle ABC = \angle ABD + \angle DBC = 31^\circ + 61^\circ = 92^\circ.

In cyclic quadrilateral ABCDABCD, ∠ABC\angle ABC and ∠ADC\angle ADC are opposite angles, so

∠ADC=180∘−∠ABC=180∘−92∘=88∘.\angle ADC = 180^\circ - \angle ABC = 180^\circ - 92^\circ = 88^\circ.

(d) angle BCDBCD

The angle at AA is

∠BAD=∠BAC+∠CAD=47∘+61∘=108∘.\angle BAD = \angle BAC + \angle CAD = 47^\circ + 61^\circ = 108^\circ.

∠BAD\angle BAD and ∠BCD\angle BCD are the other pair of opposite angles of the cyclic quadrilateral, so

∠BCD=180∘−∠BAD=180∘−108∘=72∘.\angle BCD = 180^\circ - \angle BAD = 180^\circ - 108^\circ = 72^\circ.

Check (optional)

In triangle BCDBCD the three angles should sum to 180∘180^\circ:

∠DBC+∠BCD+∠BDC=61∘+72∘+47∘=180∘.  ✓\angle DBC + \angle BCD + \angle BDC = 61^\circ + 72^\circ + 47^\circ = 180^\circ. \;\checkmark

(As a further check, in triangle ACDACD: ∠CAD+∠ACD+∠ADC\angle CAD + \angle ACD + \angle ADC; here ∠ACD=∠ABD=31∘\angle ACD = \angle ABD = 31^\circ in the same segment, giving 61∘+31∘+88∘=180∘.  ✓61^\circ + 31^\circ + 88^\circ = 180^\circ.\;\checkmark)

Final answers

  ∠BDC=47∘,∠DBC=61∘,∠ADC=88∘,∠BCD=72∘  \boxed{\;\angle BDC = 47^\circ,\quad \angle DBC = 61^\circ,\quad \angle ADC = 88^\circ,\quad \angle BCD = 72^\circ\;}