If the chord AB of a circle is parallel to the tangent at C, the prove that AC = BC

Open in App

Solution

$We have, PM ∥ AB and OC ⊥ PM at C . Hence, ∠ CQB = ∠ PCQ = 90 ° (Alternate angles) and ∠ AQO = ∠ C QB = 90 ° (Vertically opposite angle) i . e ., OQ ⊥ AB We know that a perpendicular line from the centre to the chord bisects the chord . So, AQ = QB . . . (i) In ∆ AQC and ∆ BQC, we have : AQ = QB [From equation (i)] CQ = CQ (Common) ∠ AQC = ∠ CQB = 90 ° i . e ., ∆ AQC ≅ ∆ BQC (By SAS congruency) ∴ AC = BC (CPCT)$

Suggest Corrections

Similar questions

Q. AB is a chord of a circle and CT, the tangent at C, is parallel to AB. Prove that

C A = C B

Q. AB is the diameter and AC is a chord of a circle such that

B A C = 30^{\circ}

. If tangent at C intersects AB produced in D, prove that BC=BD.

AB is diameter and AC is a chord of a circle such that $∠ B A C = 30^{\circ}$ . If the tangent at C intersects AB produced in D, prove that BC = BD [4 MARKS]

AB is a diameter and AC is chord of a circle with centre O such that $∠ B A C = 30^{\circ}$ . The tangent at C intersects AB at a point D. Prove that BC =BD.

A B

is a diameter and

A C

is a chord of a circle with centre

O

such that

∠ B A C = 30^{o}

. The tangent at

C

intersects

A B

at a point

D

. Prove that

B C = B D

Join BYJU'S Learning Program