Question

In the adjoining figure seg ab is a diameter of a circle with centre o. The bisector of angle abc intersects the circle at point d prove that seg ad is congruent seg bd
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Answer 1

Given : Segment AB is a diameter

Segment CD bisects ∠ACB

To prove : seg AD ≅ seg BD

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Construction: Draw seg OD.

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Proof: ∠ACB = 90° [Angle subtended in a semicircle]

∠DCB = ∠DCA = 45° [CD is the bisector of ∠C]

m(arcDB) = 2∠DCA = 90° [Subtended angle theorem]

∠DOB = m(arc DB) = 90° ………… (i)

[Definition of measure of arc]

segOA ≅ segOB …………. (ii)

[[Radii of the same circle] ∴ line OD is the perpendicular bisector

Now,

[From (i) and (ii)] seg AB. ∴ seg AD ≅ seg BD

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The inscribed angle theorem states that an angle $\red{θ}$inscribed in a circle is half of the central angle $\red{2θ}$that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.

Answer 2

Given : Segment AB is a diameter

Segment CD bisects ∠ACB

To prove : seg AD ≅ seg BD

____________________________________

Construction: Draw seg OD.

____________________________________

Proof: ∠ACB = 90° [Angle subtended in a semicircle]

∠DCB = ∠DCA = 45° [CD is the bisector of ∠C]

m(arcDB) = 2∠DCA = 90° [Subtended angle theorem]

∠DOB = m(arc DB) = 90° ………… (i)

[Definition of measure of arc]

segOA ≅ segOB …………. (ii)

[[Radii of the same circle] ∴ line OD is the perpendicular bisector

Now,

[From (i) and (ii)] seg AB. ∴ seg AD ≅ seg BD

___________________________________

The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.