Exercise 5.2 Practice
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Overview
This page provides comprehensive Ch 5: Circles - Exercise Set 5.2 Practice. Solve proofs showing that the triangle formed by a chord and the circle's centre is isosceles, and that equal chords form congruent central triangles with step-by-step solutions.
Isosceles Triangles & Congruence Properties of Chords
Q1: Chord and Centre Isosceles Triangle
Show that the triangle formed by a chord and the centre of the circle is isosceles.
Let $AB$ be a chord of a circle with centre $O$.
• The vertices of the triangle formed are $O$, $A$, and $B$.
• The segments $OA$ and $OB$ are radii of the circle.
• The vertices of the triangle formed are $O$, $A$, and $B$.
• The segments $OA$ and $OB$ are radii of the circle.
Since all radii of a circle are equal in length:
$$OA = OB = r$$
$$OA = OB = r$$
A triangle is classified as **isosceles** if at least two of its sides are equal in length.
Since $OA = OB$, $\Delta OAB$ is indeed an isosceles triangle.
Since $OA = OB$, $\Delta OAB$ is indeed an isosceles triangle.
Proved: OA = OB, hence ΔOAB is isosceles.
Q2: Congruence of Equal Chord Triangles
Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Let $AB$ and $CD$ be two chords of a circle with centre $O$, representing the bases of the two triangles $\Delta OAB$ and $\Delta OCD$.
• We are given that they have equal base lengths:
$$AB = CD$$
• We are given that they have equal base lengths:
$$AB = CD$$
Compare $\Delta OAB$ and $\Delta OCD$:
1. $OA = OC$ (both are radii of the circle $= r$)
2. $OB = OD$ (both are radii of the circle $= r$)
3. $AB = CD$ (given equal base length)
1. $OA = OC$ (both are radii of the circle $= r$)
2. $OB = OD$ (both are radii of the circle $= r$)
3. $AB = CD$ (given equal base length)
By the **SSS (Side-Side-Side) Congruence Criterion**, if three sides of one triangle are equal to the three corresponding sides of another triangle, the two triangles are congruent:
$$\Delta OAB \cong \Delta OCD$$
$$\Delta OAB \cong \Delta OCD$$
Proved: ΔOAB ≅ ΔOCD by SSS congruence.