Geometry of Triangles Class 9 NCERT Exercise 7.5

Core Concepts: Locus, Circumcentre, and Incentre

To solve the problems in Exercise 7.5, we must first understand the concept of a locus (plural: loci). A locus is a set of points satisfying a specific geometric condition. In a plane, two fundamental rules apply:

1. Equidistant from two points: The locus of points equidistant from two distinct points $A$ and $B$ is the perpendicular bisector of the line segment $AB$.
2. Equidistant from two intersecting lines: The locus of points equidistant from two intersecting lines is the pair of angle bisectors of the angles formed by those lines.

Applying these to a triangle gives us two remarkable points of concurrency:

• Circumcentre: The point of intersection of the perpendicular bisectors of the sides of a triangle. This point is equidistant from all three vertices ($A$, $B$, and $C$). If you draw a circle with this point as center passing through one vertex, it will pass through all three.
• Incentre: The point of intersection of the internal angle bisectors of a triangle. This point is equidistant from all three sides ($AB$, $BC$, and $CA$).

Step-by-Step Construction Methods

1. Locating a Point Equidistant from Three Vertices (Circumcentre) — 1. Draw any triangle $ABC$. 2. Construct the perpendicular bisector of side $AB$ using a compass. 3. Construct the perpendicular bisector of side $BC$. 4. Mark the point of intersection of these two bisectors as $O$. Point $O$ is the circumcentre and is equidistant from vertices $A$, $B$, and $C$ ($OA = OB = OC$).
2. Locating a Point Equidistant from Three Sides (Incentre) — 1. Draw any triangle $ABC$. 2. Construct the angle bisector of $\angle B$ using a compass. 3. Construct the angle bisector of $\angle C$. 4. Mark the point of intersection of these two angle bisectors as $I$. Point $I$ is the incentre and is equidistant from the sides $AB$, $BC$, and $AC$.

Common Exam Mistakes & Tips

Many students lose marks by confusing 'equidistant from vertices' with 'equidistant from sides'. Remember this quick rule of thumb:

• Vertices $\rightarrow$ Sides' Bisectors: If a question asks for a point equidistant from the corners/vertices (like three people standing at points A, B, and C), draw perpendicular bisectors of the sides.
• Sides $\rightarrow$ Angles' Bisectors: If a question asks for a point equidistant from the boundaries/roads/sides, draw angle bisectors.
• Always write down the steps of construction clearly in your board exams to secure full step-marks, even if your drawing is slightly imperfect.

Practice Questions with Solutions

• Q: Locate a point in the interior of $\triangle XYZ$ which is equidistant from all the vertices $X, Y, Z$. A: Step 1: Draw the triangle $XYZ$. Step 2: Construct the perpendicular bisector of side $XY$ by taking a radius greater than half of $XY$ and drawing arcs above and below the line. Step 3: Similarly, construct the perpendicular bisector of side $YZ$. Step 4: Label the intersection point of these two perpendicular bisectors as $C$. Final answer: Point $C$ is the circumcentre of $\triangle XYZ$, which is the unique point in the interior (for an acute-angled triangle) that is equidistant from vertices $X, Y$, and $Z$ ($CX = CY = CZ$).
• Q: Three friends are standing at three points $A$, $B$, and $C$ in a playground such that they form an acute-angled triangle. They want to place a ball at a point $P$ such that it is at equal distance from all three friends. Explain how to find the position of point $P$. A: Step 1: The positions of the three friends form the vertices of a triangle $ABC$. Step 2: To find a point equidistant from the vertices, we need to locate the circumcentre of $\triangle ABC$. Step 3: Draw the perpendicular bisector of segment $AB$. Step 4: Draw the perpendicular bisector of segment $BC$. Step 5: The intersection point of these two perpendicular bisectors is $P$. Final answer: Point $P$ is the circumcentre. Since $P$ lies on the perpendicular bisector of $AB$, $PA = PB$. Since $P$ lies on the perpendicular bisector of $BC$, $PB = PC$. Thus, $PA = PB = PC$, making it equidistant from all three friends.
• Q: Where will the point equidistant from all three sides of a triangle lie? Detail the construction. A: Step 1: Let the triangle be $ABC$. A point equidistant from all three sides is the incentre. Step 2: Draw the angle bisector of $\angle ABC$. Step 3: Draw the angle bisector of $\angle ACB$. Step 4: The point where these two angle bisectors meet is the Incentre, denoted by $I$. Final answer: The point is the Incentre $I$. It always lies in the interior of the triangle, regardless of whether the triangle is acute, obtuse, or right-angled.
• Q: In an equilateral triangle $ABC$, prove that the point equidistant from all vertices is the same as the point equidistant from all sides. A: Step 1: In an equilateral triangle, all three sides are equal ($AB = BC = CA$) and all three interior angles are equal ($60^\circ$). Step 2: The perpendicular bisectors of the sides of an equilateral triangle also act as the angle bisectors of the opposite vertices. Step 3: Since the perpendicular bisectors (which define the circumcentre) and the angle bisectors (which define the incentre) are the same lines, their point of intersection must be identical. Final answer: Therefore, in an equilateral triangle, the circumcentre and the incentre coincide at the same point.

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