Giving brainliest! Please give me the working out for the answer

∠BCD is a right angle ⇒ Proved down

Step-by-step explanation:

Let us revise some fact in the circle

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

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation:

(-5,-28)

i graphed to on desmos.