Consider the given fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH Class 9

Consider the given fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH Class 9

Question 1.

Consider the given fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF. Class 9

Answer:

Chords of a circle that are equidistant from the centre have equal length.

Given: C is the centre of the circle with radius r.

CE ⊥ AB, CH ⊥ FG and CE = CH.

To show: AB = GF

Proof:

In ∆CEA and ∆CHF,

CA = CF ... [Radii of the same circle]

CE = CH ...[Given]

∠CEA = ∠CHF = 90° ... [CE ⊥ AB, CH ⊥ FG]

∴ ∆CEA ≅ ∆CHF ... [By RHS congruence]

∴ AE = FH ... (i) [Corresponding parts of congruent triangles]

E is the midpoint of AB and H is the midpoint of GF.

... [The perpendicular from the centre to a chord bisects the chord]

∴ AB = 2AE and GF = 2FH ... (ii)

∴ AB = GF .......[From (i) and (ii)]