NCERT Solutions for Class 9th Maths Chapter 5 Exercise Set 5.2 — Chords and the Angles They Subtend
Book page 100 Updated on2026-09-08
Q1.
Show that the triangle formed by a chord and the centre of the circle is isosceles.
Answer
Given: a circle with centre O and a chord AB. Join OA and OB. To show: ΔOAB is isosceles.
OA = radius of the circle
OB = radius of the circle
∴ OA = OB
A triangle with two equal sides is isosceles, so ΔOAB is isosceles with base AB.
Consequence used again and again in this chapter: the base angles are equal, ∠OAB = ∠OBA.
Two sides of ΔOAB are radii, so they are equal — the triangle is isosceles.
Why this is worth stating: every proof about chords in this chapter starts by joining the ends of the chord to the centre. The instant you do that you own two equal sides for free, and isosceles triangles hand you equal base angles. That is the engine behind Theorems 2, 3, 4 and 9.
Q2.
Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Answer
Given: chords AB and DE of the same circle, centre C, with AB = DE. To show: ΔCAB ≅ ΔCDE.
CA = CB = r (radii)
CD = CE = r (radii)
⇒ CA = CD and CB = CE
AB = DE (given — the two bases are equal)
By the SSS congruence rule, ΔCAB ≅ ΔCDE
What follows at once: corresponding parts of congruent triangles are equal, so
∠ACB = ∠DCE — equal chords subtend equal angles at the centre (Theorem 2)
and the altitudes from C are equal — equal chords are equidistant from the centre (Theorem 6)
Tip: if instead the two triangles have equal apex angles at C, use SAS (two radii and the included angle) and you get AB = DE — that is Theorem 3, the converse.