Let and be two points on a circle with centre . (i) Are there points on the circle, on the same side of , such that is different from ? (ii) Is it true that if , then and lie on the same side of the circle? (iii) If , and and do not lie on the circle, does the circle through , and also pass through ?

Understand the Question Angles in the same segment: Any two angles subtended by the same chord at points on the same segment (or arc) of a circle are equal (). Cyclic quadrilateral property: Opposite angles of a cyclic quadrilateral sum to . Converse theorem (Concyclicity): If a line segment joining two points subtends equal angles at two other points lying on the same side of the line, then all four points lie on a circle (they are concyclic). Part (i) Are there points on the circle, on the same side of , such that is different from ? Step 1 · Analyze Angles Subtended in the Same Segment By the theorem on angles in a circle, angles subtended by the same chord in the same segment are equal: Since and lie on the same side of , they are in the same segment. Therefore, and must always be equal. Answer > (i) No Part (ii) Is it true that if , then and lie on the same side of the circle? Step 1 · Check the Counterexample with a Diameter Consider and on opposite sides of chord on the circle. Quadrilateral is cyclic, so its opposite angles are supplementary: If : When is a diameter, , even though and lie on opposite sides of . Answer > (ii) No, not always true Part (iii) If , and and do not lie on the circle, does the circle through , and also pass through ? Step 1 · Apply the Converse of the Same-Segment Theorem By the converse of the theorem for angles in the same segment: If a line segment subtends equal angles at two points and lying on the same side of the line containing the segment: then the four points are concyclic. Assuming and lie on the same side of , the circle passing through must also pass through . Answer > (iii) Yes Common Mistakes Assuming points must be on the same side: In part (ii), forgetting the case where is a diameter, which makes both angles even when and are on opposite sides of . Missing the 'same side' condition: The converse theorem for concyclicity in part (iii) strictly requires points and to lie on the same side of line segment .

Circles and Geometric Shapes | Exercise 5.6

Question 2

Let $A$ and $B$ be two points on a circle with centre $O$ .

(i) Are there points $X, Y$ on the circle, on the same side of $AB$ , such that $\angle AXB$ is different from $\angle AYB$ ?

(ii) Is it true that if $\angle AXB = \angle AYB$ , then $X$ and $Y$ lie on the same side of the circle?

(iii) If $\angle AXB = \angle AYB$ , and $X$ and $Y$ do not lie on the circle, does the circle through $A$ , $B$ and $X$ also pass through $Y$ ?

Solution

Understand the Question

Angles in the same segment: Any two angles subtended by the same chord at points on the same segment (or arc) of a circle are equal ( $\angle AXB = \angle AYB$ ).
Cyclic quadrilateral property: Opposite angles of a cyclic quadrilateral sum to $180^\circ$ .
Converse theorem (Concyclicity): If a line segment joining two points subtends equal angles at two other points lying on the same side of the line, then all four points lie on a circle (they are concyclic).

(i) Are there points $X, Y$ on the circle, on the same side of $AB$ , such that $\angle AXB$ is different from $\angle AYB$ ?

Step 1 · Analyze Angles Subtended in the Same Segment

Diagram 1

By the theorem on angles in a circle, angles subtended by the same chord $AB$ in the same segment are equal: $\angle AXB = \angle AYB$

Since $X$ and $Y$ lie on the same side of $AB$ , they are in the same segment. Therefore, $\angle AXB$ and $\angle AYB$ must always be equal.

Answer

(i) No

(ii) Is it true that if $\angle AXB = \angle AYB$ , then $X$ and $Y$ lie on the same side of the circle?

Step 1 · Check the Counterexample with a Diameter

Diagram 2

Consider $X$ and $Y$ on opposite sides of chord $AB$ on the circle. Quadrilateral $AXBY$ is cyclic, so its opposite angles are supplementary: $\angle AXB + \angle AYB = 180^\circ$

If $\angle AXB = \angle AYB$ :

\begin{aligned} 2\angle AXB &= 180^\circ \\[0.4em] \angle AXB &= 90^\circ \end{aligned}

When $AB$ is a diameter, $\angle AXB = \angle AYB = 90^\circ$ , even though $X$ and $Y$ lie on opposite sides of $AB$ .

Answer

(ii) No, not always true

(iii) If $\angle AXB = \angle AYB$ , and $X$ and $Y$ do not lie on the circle, does the circle through $A$ , $B$ and $X$ also pass through $Y$ ?

Step 1 · Apply the Converse of the Same-Segment Theorem

Diagram 3

By the converse of the theorem for angles in the same segment: If a line segment $AB$ subtends equal angles at two points $X$ and $Y$ lying on the same side of the line containing the segment: $\angle AXB = \angle AYB$ then the four points $A, B, X, Y$ are concyclic.

Assuming $X$ and $Y$ lie on the same side of $AB$ , the circle passing through $A, B, X$ must also pass through $Y$ .

Answer

(iii) Yes

Common Mistakes

Assuming points must be on the same side: In part (ii), forgetting the case where $AB$ is a diameter, which makes both angles $90^\circ$ even when $X$ and $Y$ are on opposite sides of $AB$ .
Missing the 'same side' condition: The converse theorem for concyclicity in part (iii) strictly requires points $X$ and $Y$ to lie on the same side of line segment $AB$ .

Question 2

More questions in Exercise 5.6