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Theorem: A straight line that joins the centre, O, of a circle to the midpoint of a chord, which is not a diameter, is perpendicular to the cord.

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Given: A circle with centre O and chord AB. M is joined to centre which divides the chord AB such that AM = MB.

To prove: OM is perpendicular to AB.

Proof:

OA = OB (radii of the same circle)

OM = (common side)

∴ △AMO ≅ △BMO

Screenshot 2022 06 05 at 23.22.46

∠AMO + ∠BMO = 180º

but ∠AMO = ∠BMO = \( \frac{180}{2} \\ \scriptsize 90^o\)

Therefore, OM is perpendicular to AB.

The convex of the Theorem above is also true:

The perpendicular bisector of a chord passes through the centre of the circle.

Example 1:

In the diagram below, 𝐴𝑀 = 200cm and 𝑀𝐶 = 120cm, find 𝐴𝐵.

Screenshot 2022 06 06 at 11.53.05

Solution:

Given:

Circle wth centre M

Chord = AB

MD bisects AB at C

∴ ∠MCB = 90º

AM = 200cm

AM is a radius of the circle

BM is also a radius of the circle

∴ BM = 200cm

MC = 120cm

Let’s sketch the diagram with our given information.

Screenshot 2022 06 06 at 12.04.38

Consider △MCB

We can find BC using Pythagorean theorem.

⇒ \( \scriptsize MB^2 = MC^2 \: + \: BC^2 \)

⇒ \( \scriptsize BC^2 = MB^2 \: – \: MC^2 \)

⇒ \( \scriptsize BC = \sqrt{MB^2 \: – \: MC^2} \)

⇒ \( \scriptsize BC = \sqrt{200^2 \: – \: 120^2} \)

⇒ \( \scriptsize BC = 160 \: cm \)

MC bisects AB

so AB = 2 x BC

∴ AB = 2 x 160cm

AB = 320cm

Example 2:

\( \scriptsize \overline{AB} \) and \( \scriptsize \overline{AC} \) are two chords in a circle on opposite sides of center 𝑀, where ∠BAC =33º. If 𝐷 and 𝐸 are the midpoints of \( \scriptsize \overline{AB} \) and \( \scriptsize \overline{AC} \), find ∠DME.

Screenshot 2022 06 07 at 00.32.17

Solution:

From the diagram \( \scriptsize \overline{ME} \) and \( \scriptsize \overline{MD} \) pass through the centre and bisect \( \scriptsize \overline{AB} \) and \( \scriptsize \overline{AC} \)

∴ ∠MEA = 90º (perpendicular line to mid-point of chord)

∠MDA = 90º (perpendicular line to mid-point of chord)

We can redraw the diagram with the new information.

Screenshot 2022 06 07 at 00.51.12

Let us consider quadrilateral ADME

∠MEA = 90º
∠MDA = 90º
∠DAE = 33º
∠DME = ?

Since the interior angles of a quadrilateral always add up to 360º, we can find the measure of the unknown angle as follows:

90º + 90º + 33º + ∠DME = 360º

∠DME = 360 – 90 – 90 – 33

∠DME = 147º

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