Question

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that
angleCBD + angleCDB = 1/2 BAD.​

Answer 1

Answer:

• Hence Proved!!

Step-by-step explanation:

Given:

• OBDC is a quadrilateral.
• O is the centre of the circle passing through points B, C and D.

To Prove:

• ∠DCB + ∠CBD = 1/2 ∠BOC

Note:

• Diagram attached in the attachment file.

Now, we know that ∠ substended by chord at centre of circle is double the angle substended by it at the remaining part of circle.

So, ∠BOC = 2∠BAC

And, we know that ABDC is cyclic quadrilateral.

=> ∠BAC + ∠BDC = 180° [Opposite angle of cyclic quadrilateral is 180°]

=> ∠BAC = 180° - ∠BDC

=> ∠BOC = 360° - 2∠BDC

=> 2∠BDC = 360° - ∠BOC

=> ∠BDC = (360° - ∠BOC)/2

=> ∠BDC = 180° - (∠BOC/2)

Now, In triangle BDC,

By angle sum property,

=> ∠BDC + ∠DCB + ∠CBD = 180°

=> [180° - (∠BOC/2)] + ∠DCB + ∠CBD = 180°

=> ∠DCB + ∠CBD = 180° - [180° - (∠BOC/2)]

=> ∠DCB + ∠CBD = 180° - 180° + (∠BOC/2)

=> ∠DCB + ∠CBD = 1/2 ∠BOC

Hence Proved!!

Answer 2

$\huge \mathfrak \red{answer}$

ABCD is a quadrilateral

$\sf{A \:is \: centre \: of \: circle}$

$\sf{BD \:is \: a \: chord}$

$\sf{∠BAD \: = 2 \: ∠BED}</p><p>$

$\sf{EBCD \: is \: cyclic \: quadrilateral}$

$\sf{∠BED \: + \: ∠BCD = 180}</p><p></p><p>$

$\sf{∠BED = 180 - ∠BCD}</p><p></p><p>$

$\sf{∠BAD = 2∠BED \: \: 360 - 2∠BCD}</p><p></p><p>$

$\rm{∠BCD = \frac{360 -∠BAD</p><p>}{2} \: 180 - \frac{ ∠BAD }{2} }</p><p>$

$\sf \red{9n \: triangle \: BCD}</p><p></p><p></p><p>$

$\sf \red{ ∠CBD +∠CBD + ∠BCD = 180}$

$\sf{ ∠CBD + ∠CDB = 180 - ∠BCD}$

$\tt{180 - (180 - \frac{∠bad }{2})}$

$\tt \red{180 - 180 + \frac{∠BAD}{2}}$

$\sf{ \frac{∠BAD}{2}}$

I hope it's help uh