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Chords and their perpendicular bisectors · 5 marks

In a circle, two chords CC’ and DD’ are drawn perpendicular to a diameter AB. Prove that the segment MM’ joining the midpoints of the chords CD and C’ D’ is perpendicular to AB.

Answer: The diameter AB is perpendicular to CC′ and DD′, so it bisects them: C′ and D′ are the mirror images of C and D in AB. Using congruent triangles (or this reflection), CD and C′D′ are mirror images, so their midpoints M and M′ are mirror images too, and MM′ ⟂ AB.

Step-by-step solution

Given: AB is a diameter; Chords CC′ ⟂ AB at P and DD′ ⟂ AB at Q; M, M′ are the midpoints of CD and C′D′
To find: Prove MM′ ⟂ AB

Idea: A diameter perpendicular to a chord bisects it, so AB is the perpendicular bisector of CC′ and of DD′: the whole figure is symmetric about AB. A chain of congruent triangles turns this symmetry into a proof.

ABCC′DD′MM′PQR
  1. Let CC′ meet AB at P and DD′ meet AB at Q, with C, D on one side of AB. AB passes through the centre and is perpendicular to both chords, so it bisects them: CP = PC′ and DQ = QD′. (If C and D lie on opposite sides of AB, the reflection argument in the last step proves the result directly.)1 mark
  2. ΔCPQ ≅ ΔC′PQ (SAS: CP = C′P, ∠CPQ = ∠C′PQ = 90°, PQ common). So CQ = C′Q and ∠CQP = ∠C′QP; hence ∠CQD = ∠C′QD′ (each is the difference between a right angle and these equal angles).1 mark
  3. ΔCQD ≅ ΔC′QD′ (SAS: CQ = C′Q, ∠CQD = ∠C′QD′, QD = QD′). So CD = C′D′ and ∠QDC = ∠QD′C′.1 mark
  4. DM = ½CD = ½C′D′ = D′M′. So ΔQDM ≅ ΔQD′M′ (SAS: QD = QD′, ∠QDM = ∠QD′M′, DM = D′M′), giving QM = QM′ and ∠DQM = ∠D′QM′; then ∠MQR = ∠M′QR (each is 90° minus these equal angles), where R is the point where MM′ meets AB.1 mark
  5. ΔQRM ≅ ΔQRM′ (SAS: QM = QM′, ∠MQR = ∠M′QR, QR common), so ∠QRM = ∠QRM′. These form a linear pair, so each is 90°: MM′ ⟂ AB.1 mark
  6. In one line: AB is the perpendicular bisector of CC′ and DD′, so reflecting in AB swaps C with C′ and D with D′, carries CD onto C′D′ and M onto M′; a point and its mirror image are joined by a segment perpendicular to the mirror.
Because the diameter AB bisects the perpendicular chords CC′ and DD′, the figure is symmetric about AB; the congruent triangles CPQ, CQD, QDM and QRM with their mirror partners give ∠QRM = ∠QRM′ = 90°. Hence MM′ is perpendicular to AB.

Answer to write in the exam

Let CC′ ⟂ AB at P, DD′ ⟂ AB at Q; MM′ meets AB at R.

CP = PC′, DQ = QD′ (diameter ⟂ chord bisects the chord)

ΔCPQ ≅ ΔC′PQ (SAS) ⇒ CQ = C′Q, ∠CQP = ∠C′QP ⇒ ∠CQD = ∠C′QD′

ΔCQD ≅ ΔC′QD′ (SAS) ⇒ CD = C′D′, ∠QDC = ∠QD′C′

DM = ½CD = ½C′D′ = D′M′ ⇒ ΔQDM ≅ ΔQD′M′ (SAS) ⇒ QM = QM′, ∠DQM = ∠D′QM′ ⇒ ∠MQR = ∠M′QR

ΔQRM ≅ ΔQRM′ (SAS: QM = QM′, ∠MQR = ∠M′QR, QR common) ⇒ ∠QRM = ∠QRM′

∠QRM + ∠QRM′ = 180° (linear pair) ⇒ ∠QRM = 90°

∴ MM′ ⟂ AB

Common mistakes that cost marks

  • Assuming CD ∥ C′D′ or that CD is perpendicular to AB; neither is given.
  • Forgetting to state why P and Q are midpoints: a diameter perpendicular to a chord bisects it.
  • Ending at QM = QM′ without showing the angle at R is 90°.

How this can come in the exam

MCQ (1 mark)

A diameter AB of a circle is perpendicular to a chord CC′ of length 10 cm and meets it at P. Then CP is

  1. 2.5 cm
  2. 5 cm
  3. 10 cm
  4. 20 cm
Show answer

(B) 5 cm
A diameter perpendicular to a chord bisects it, so CP = 10 ÷ 2 = 5 cm.

Try one yourself

In the same figure, show that CD = C′D′.

Show answer

CP = PC′ and DQ = QD′ (the diameter bisects the perpendicular chords). ΔCPQ ≅ ΔC′PQ (SAS) gives CQ = C′Q and ∠CQD = ∠C′QD′; then ΔCQD ≅ ΔC′QD′ (SAS), so CD = C′D′.

More questions like this

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