Quadrilateral ABCD is inscribed in circle Pas shown. Which statement is necessarily true?
A MZA+ m2B=m2C+ m2D
B. MZA+ m2C=mZB+ m2D
C MZA+m2D=mZB+mzo
D. MZA+mZB = 2(m2C+ m2D)

Given:

Quadrilateral ABCD is inscribed in a circle P.

To find:

Which statement is necessarily true.

Solution:

Quadrilateral ABCD is inscribed in a circle P.

Therefore ABCD is a cyclic quadrilateral.

In cyclic quadrilateral, opposite angles form a supplementary angles.

⇒ m∠A + m∠C = 180° (1)

⇒ m∠B + m∠D = 180° (2)

By (1) and (2),

⇒ m∠A + m∠C = m∠B + m∠D

This statement is necessarily true for the quadrilateral ABCD in circle P.

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: