Answer:
∠BCD is a right angle ⇒ Proved down
Step-by-step explanation:
Let us revise some fact in the circle
- If the vertices of a quadrilateral lie on a circumference of a circle, then this quadrilateral is called a cyclic quadrilateral
- In a cyclic quadrilateral, every two opposite angles are supplementary (the sum of their measures is 180°)
Let us use these facts to solve the question
∵ ABCD is a quadrilateral
∵ Its vertices lie of the circumference of the circle
∴ ABCD is a cyclic quadrilateral
By using the facts above
∴ ∠ABC and ∠ADC are supplementary
∴ m∠ABC + m∠ADC = 180° → (1)
∵ m∠ABD = m∠CBD → given
∵ m∠ABC = m∠ABD + m∠CBD
∴ m∠ABC = 2m∠CBD → (2)
∵ m∠ADB = m∠CDB → given
∵ m∠ADC = m∠ADB + m∠CDB
∴ m∠ADC = 2m∠CDB → (3)
→ Substitute (2) and (3) in (1)
∴ 2m∠CBD + 2m∠CDB = 180
→ Divide both sides by 2
∴ m∠CBD + m∠CDB = 90
In ΔBCD
∵ m∠BCD + m∠CBD + m∠CDB = 180 → Sum of angles of a triangle
∵ m∠CBD + m∠CDB = 90
∴ m∠BCD + 90 = 180
→ Subtract 90 from both sides
∴ m∠BCD + 90 - 90 = 180 - 90
∴ m∠BCD = 90°
∴ ∠BCD is a right angle ⇒ proved
