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प्रश्न
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate : ∠BDC
उत्तर
Given that BD is a diameter of the circle.
The angle in a semicircle is a right angle.
∴ ∠ BCD = 90°
Also given that ∠ DBC = 58°
In Δ BDC ,
∠ BCD + ∠ BCD + ∠ BDC = 180°
⇒ 58° + 90° + ∠ BDC = 180°
⇒ 148° + ∠ BDC = 180°
⇒ ∠ BDC = 180° - 148°
⇒ ∠ BDC = 32°
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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
