Question

In figure quadrilateral ABCD is cyclic , if m(arc BC) = 90° and angle DBC = 55° then find measure of angle BCD​

Answer 1

m(arc BC)=90° ⇒(given)

∴m∠BDC= \begin{gathered}\frac{1}{2\\}\end{gathered} m(arc BC) ⇒(Inscribed Angle Theorem)

∴m∠BDC= \begin{gathered}\frac{1}{2\\}\end{gathered} ×90°

∴m∠BDC=45° ⇒(1)

Now, Consider ΔBDC

m∠BDC + m∠BCD + m∠DBC = 180° ⇒( Angle sum property of Triangle )

But, m∠BDC=45° ⇒(from 1)

m∠DBC = 55° ⇒ (given)

∴ 45° + m∠BCD + 55° = 180°

∴ m∠DCB + 100° = 180°

∴ m∠BCD = 180° - 100°

∴ m∠BCD = 80°

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Answer 2

Given : quadrilateral ABCD is cyclic ,  m(arc BC) = 90° and angle DBC = 55°  

To find : measure of angle BCD​

Solution:

Let say O is the center of circle passing through quadrilateral vertex

m(arc BC) = 90°

=> ∠BOC = 90°

in Δ BOC

OB = OC = Radius

=> ∠OBC = ∠OCB

∠OBC + ∠OCB + ∠BOC = 180°   ( Sum of angles of triangle)

=> ∠OBC = ∠OCB  = 45°

∠DBC = 55°

=>  ∠DOC = 2∠DBC  = 2 x 55°  = 110°

in Δ DOC

OC = OD = Radius

=> ∠OCD = ∠ODC  = 35°    ( (180 - 110)/2 )

∠BCD =   ∠OCB +  ∠OCD

= 45° + 35°

= 80°

∠BCD =  80°

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