Question

If AOB is a diameter of a circle and CD ║ AB if ∠BAD = 30° THEN ∠CAD = ?

Answer 1

BAD = ADC
= 30

BAD = BCD
= 30

ACB =90
ACD = ACB + BCD
= 120

ACD + CDA +DAC = 180
120 + 30 + CAD =180
CAD = 30

Answer 2

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answer:

Given: CD||AB and ∠BAD = 30°

Consider ΔABD

∠ADB = 90° (angle in semicircle)

Now, by angle sum property

∠ABD + ∠BAD + ∠ADB = 180°

∠ABD + 30° + 90° = 180°

∠ABD = 180° – 30° – 90°

∠ABD = 60°

Here,

∠ABD + ∠ACD = 180° (opposite angles in cyclic quadrilateral are supplementary)

60° + ∠ACD = 180°

∠BCD = 180° – 60°

∠BCD = 120°

Here, CD||AB and AC is the transversal

∠CAB + ∠ACD = 180° (interior angles along the transversal are supplementary)

∠CAB + 120° = 180°

∠ABC = 180° – 120° = 60°

∠ABC = 60°

∠ABC = ∠CAD + ∠DAB

60° = ∠CAD + 30°

∠CAD = 60° – 30° = 30°

∴ ∠CAD = 30°Given: CD||AB and ∠BAD = 30°

Consider ΔABD

∠ADB = 90° (angle in semicircle)

Now, by angle sum property

∠ABD + ∠BAD + ∠ADB = 180°

∠ABD + 30° + 90° = 180°

∠ABD = 180° – 30° – 90°

∠ABD = 60°

Here,

∠ABD + ∠ACD = 180° (opposite angles in cyclic quadrilateral are supplementary)

60° + ∠ACD = 180°

∠BCD = 180° – 60°

∠BCD = 120°

Here, CD||AB and AC is the transversal

∠CAB + ∠ACD = 180° (interior angles along the transversal are supplementary)

∠CAB + 120° = 180°

∠ABC = 180° – 120° = 60°

∠ABC = 60°

∠ABC = ∠CAD + ∠DAB

60° = ∠CAD + 30°

∠CAD = 60° – 30° = 30°

∴ ∠CAD = 30°