Consider the figure. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
Step-by-step solution
To find: Show that AB = GF
Idea: This is the converse of “equal chords are equidistant from the centre”: chords equidistant from the centre are equal. The perpendicular from the centre bisects a chord, so it is enough to show the half-chords AE and FH are equal.
- Join CA and CF. In ΔCEA and ΔCHF: ∠CEA = ∠CHF = 90° (given).½ mark
- Hypotenuse CA = hypotenuse CF (radii) and CE = CH (given).½ mark
- So ΔCEA ≅ ΔCHF (RHS), and AE = FH (CPCT).1 mark
- The perpendicular from the centre to a chord bisects the chord, so AB = 2AE and GF = 2FH.½ mark
- Hence AB = 2AE = 2FH = GF. Chords equidistant from the centre are equal.½ mark
Answer to write in the exam
Given: CE ⟂ AB, CH ⟂ GF, CE = CH. To prove: AB = GF.
In ΔCEA and ΔCHF:
∠CEA = ∠CHF = 90° (given)
CA = CF (radii)
CE = CH (given)
∴ ΔCEA ≅ ΔCHF (RHS) ⇒ AE = FH (CPCT)
AB = 2AE, GF = 2FH (perpendicular from the centre bisects the chord)
∴ AB = GF
Common mistakes that cost marks
- Stopping at AE = FH. The question asks for the whole chords, so the bisecting step is needed.
- Using SAS with CA = CF and CE = CH but the angle at C, which is not known to be equal.
- Writing CH ⟂ GH and thinking H is outside the chord. H lies on GF, so GH is part of GF.
How this can come in the exam
Chords AB and CD of a circle are each 4 cm from the centre, and AB = 7 cm. Then CD is
- 4 cm
- 3.5 cm
- 7 cm
- 14 cm
Show answer
(C) 7 cm
Chords equidistant from the centre are equal, so CD = AB = 7 cm.
Try one yourself
Two chords of a circle of radius 26 cm are each 10 cm from the centre. Find the length of each chord.
Show answer
Half-chord = √(262 − 102) = √576 = 24 cm, so each chord = 48 cm; they are equal, as expected.
More questions like this
- If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF. Solve this using the Baudhāyana–Pythagoras theorem.
- You have two chords on a circle. One is longer than the other. Which chord is closer to the centre? Can you guess?
Activity: Draw a circle. Draw chords of various lengths. Drop a perpendicular to each chord from the centre. Record the length of the chord and its distance from the centre in a table.
What do you observe? - Let AB and DE be two chords of a circle with centre C. Suppose AB > DE. Then the distance from C to AB is less than the distance from C to DE.
- Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
- Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r2 − d2).