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प्रश्न
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
उत्तर
∠ADB = ∠ACB ...(i) (Angles in same segement)
Similarly,
∠ABD = ∠ACD ...(ii)
But, ∠ACB = ∠ACD ...(AC is bisector of ∠BCD)
∴ ∠ADB = ∠ABD ...(From (i) and (ii))
TAS is a tangent and AB is a chord
∴ ∠BAS = ∠ADB ...(Angles in alternate segment)
But, ∠ADB = ∠ABD
∴ ∠BAS = ∠ABD
But these are alternate angles
Therefore, TS || BD
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संबंधित प्रश्न
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In cyclic quadrilateral ABCD, ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°. Calculate : ∠DCB.
In the given below figure,
∠ BAD = 65°
∠ ABD = 70°
∠ BDC = 45°
Find: (i) ∠ BCD, (ii) ∠ ADB.
Hence show that AC is a diameter.
An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle, to prove the theorem complete the activity.
Given: ABCD is cyclic,
`square` is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
Proof: `square` + ∠BCD = `square` .....[Angles in linear pair] (I)
ABCD is a cyclic.
`square` + ∠BAD = `square` ......[Theorem of cyclic quadrilateral] (II)
By (I) and (II)
∠DCE + ∠BCD = `square` + ∠BAD
∠DCE ≅ ∠BAD
