AB is the diameter and AC is a chord of a circle such that BAC -30- If tangent at C intersects AB produced in D- prove that BC - BD-

∠ DCB = ∠ BAC = 30^∘ (angle bet tangent amp; chord BC = angle the opp segment) ∠ ACB = 90^∘ (angle is semicircle) ∴∠ CBA = 60^∘ (from Δ ABC ∠ CBA = ∠ BCD + ...

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Standard X

Mathematics

Relations Using Incircle and Outcircle

AB is the dia...

Question

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.

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Solution

$∠ D C B = ∠ B A C = 30^{\circ}$ (angle bet tangent & chord BC = angle the opp segment)
$∠ A C B = 90^{\circ}$ (angle is semicircle)
$∴ ∠ C B A = 60^{\circ} (f r o m Δ A B C ∠ C B A = ∠ B C D + ∠ C D B 60^{\circ} = 30^{\circ} + x \Rightarrow ∠ C D B = 30^{\circ} ∴ Δ C B D i s i s o c e l l s ∠ B C D = ∠ C D B = 30^{\circ}$
$\Rightarrow B C = B D$ Proved

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AB is the diameter and AC is a chord of a circle such that BAC=30∘ . If tangent at C intersects AB produced in D, prove that BC=BD.

∠DCB=∠BAC=30∘ (angle bet tangent & chord BC = angle the opp segment) ∠ACB=90∘ (angle is semicircle) ∴∠CBA=60∘(fromΔABC∠CBA=∠BCD+∠CDB60∘=30∘+x⇒∠CDB=30∘∴ΔCBDis isocells∠BCD=∠CDB=30∘ ⇒BC=BD Proved

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.