Question 2

We can find different triangles by connecting points on the circles.

Step 1 — Understanding the circle properties

Let us call the radius of each circle $r$. All circles shown have this same radius $r$. In the first diagram, circle A has its center at A. It passes through point B. So, the distance from A to B is $r$. We write this as $AB = r$. Circle B has its center at B. It passes through point A. So, the distance from B to A is also $r$. In the second diagram, circle A passes through B and C. So, $AB = r$ and $AC = r$. Circle B passes through A and C. So, $BA = r$ and $BC = r$. Circle C passes through A and B. So, $CA = r$ and $CB = r$. This means the distances $AB$, $BC$, and $CA$ are all equal to $r$.

Step 2 — Forming isosceles triangles

An isosceles triangle has two sides of equal length. Let us look at the first diagram. The two circles intersect at points X and Y. Consider triangle AXY. Side $AX$ is a radius of circle A. So, $AX = r$. Side $AY$ is a radius of circle A. So, $AY = r$. Since $AX = AY$, triangle AXY has two equal sides. So, triangle AXY is an isosceles triangle. Now consider triangle BXY. Side $BX$ is a radius of circle B. So, $BX = r$. Side $BY$ is a radius of circle B. So, $BY = r$. Since $BX = BY$, triangle BXY has two equal sides. So, triangle BXY is an isosceles triangle.

Step 3 — Forming equilateral triangles

An equilateral triangle has all three sides of equal length. Let us look at the first diagram again. Consider triangle AXB. Side $AX$ is a radius of circle A. So, $AX = r$. Side $BX$ is a radius of circle B. So, $BX = r$. Side $AB$ is the distance between centers. We found $AB = r$. All three sides are equal: $AX = BX = AB = r$. So, triangle AXB is an equilateral triangle. Similarly, consider triangle AYB. Side $AY$ is a radius of circle A. So, $AY = r$. Side $BY$ is a radius of circle B. So, $BY = r$. Side $AB$ is the distance between centers. We found $AB = r$. All three sides are equal: $AY = BY = AB = r$. So, triangle AYB is an equilateral triangle. Now let us look at the second diagram. Consider triangle ABC. Side $AB$ is the distance between centers. We found $AB = r$. Side $BC$ is the distance between centers. We found $BC = r$. Side $CA$ is the distance between centers. We found $CA = r$. All three sides are equal: $AB = BC = CA = r$. So, triangle ABC is an equilateral triangle. Let P be the intersection of circle A and circle B, not C. Consider triangle PAB. Side $PA$ is a radius of circle A. So, $PA = r$. Side $PB$ is a radius of circle B. So, $PB = r$. Side $AB$ is the distance between centers. We found $AB = r$. All three sides are equal: $PA = PB = AB = r$. So, triangle PAB is an equilateral triangle. Let Q be the intersection of circle B and circle C, not A. Consider triangle QBC. Side $QB$ is a radius of circle B. So, $QB = r$. Side $QC$ is a radius of circle C. So, $QC = r$. Side $BC$ is the distance between centers. We found $BC = r$. All three sides are equal: $QB = QC = BC = r$. So, triangle QBC is an equilateral triangle. Let R be the intersection of circle C and circle A, not B. Consider triangle RAC. Side $RA$ is a radius of circle A. So, $RA = r$. Side $RC$ is a radius of circle C. So, $RC = r$. Side $AC$ is the distance between centers. We found $AC = r$. All three sides are equal: $RA = RC = AC = r$. So, triangle RAC is an equilateral triangle.

Step 4 — Equilateral triangles are also isosceles

An equilateral triangle has all three sides equal. An isosceles triangle needs only two sides to be equal. Since all three sides are equal in an equilateral triangle, any two sides are equal. Therefore, every equilateral triangle is also an isosceles triangle.

Answer

(i) Isosceles triangles are AXY and BXY. (ii) Equilateral triangles from two circles are AXB and AYB. (iii) Equilateral triangles from three circles are ABC, PAB, QBC, and RAC. All these equilateral triangles are also isosceles triangles.