ESTION 2

Question

2.1 In the diagram below, 𝑂 is the centre of the circle and 𝐴 and 𝐵 are
points on the circumference of the circle. 𝑀 is a point on chord 𝐴𝐵
such that 𝑂𝑀 ⊥ 𝐴𝐵.
Figure 3: A circle with centre 𝑂.
Prove the theorem that states that if 𝑂𝑀 ⊥ 𝐴𝐵 then 𝐴𝑀 = 𝑀𝐵.
Redraw the diagram above to show any necessary constructions. (6)

Answer 1

To prove that if $OM\perp AB$ (where $O$ is the center of the circle, $A$ and $B$ are points on the circumference, and $M$ is a point on chord $AB$), then $AM=MB$, we can follow these steps:

Step-by-step proof:

1. Construct Radius Lines: Draw the radii $OA$ and $OB$. Since $A$ and $B$ lie on the circumference, both lines are equal in length (as they are radii of the same circle): $$OA=OB=r(where,r,is,the,radius,of,the,circle)$$

2. Right Triangle Formed: Since $OM\perp AB$, we have two right triangles: $OMA$ and $OMB$.

3. Use Right Triangle Properties: In triangles $OMA$ and $OMB$:

   • Both triangles share the side $OM$.
   • $OA$ and $OB$ are both radii of the circle and thus equal.

4. Applying the Pythagorean Theorem: According to the Pythagorean theorem: $$O{A}^2=O{M}^2+A{M}^2\text{and}O{B}^2=O{M}^2+M{B}^2$$ Substituting $OA=OB$: $$O{A}^2=O{B}^2⟹O{M}^2+A{M}^2=O{M}^2+M{B}^2$$

5. Simplify the Equation: By subtracting $O{M}^2$ from both sides, we arrive at: $$A{M}^2=M{B}^2$$

6. Conclusion: Taking the square root of both sides gives: $$AM=MB$$ Therefore, we have proven that if $OM\perp AB$, then $AM=MB$.

Diagram:

Here's a simple representation of the diagram you can sketch to illustrate the proof:

   B
  / \
 /   \
/     \
O------M
\     /
 \   /
  \ /
   A

• The point $O$ is the center of the circle.
• $A$ and $B$ are points on the circumference.
• $M$ is the midpoint of chord $AB$, where $OM\perp AB$.
• The line segments $OA$ and $OB$ are equal, and $AM$ is equal to $MB$.

Summary:

Thus, the proof demonstrates that if $OM$ is perpendicular to chord $AB$, the segment $M$ divides the chord $AB$ into two equal parts, establishing $AM=MB$.