Let the measure of BCD = aº Because BCD and BAD form a circle, and a circle measures 360°, the measure of BAD is 360 -
a”. Because of the theorem, mzA= degrees and m C = 360 -2 degrees. The sum of the measures of
angles A and Cis+350 – a degrees, which is equal to 360°, or 180°. Therefore, angles A and C are supplementary
because their measures add up to 180° Angles B and D are supplementary because the sum of the measures of the
angles in a quadrilateral is 360° MZA+ m2C + m2B + m2D = 360° and using substitution, 180° + m2B + m2D = 360°
so mZB + m2D = 180°
What is the missing information in the paragraph proof?
inscribed angle
polygon interior angle sum
quadrilateral angle sum
angle bisector

where are the choices?

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.

n word

step-by-step explanation:

what lives in the hood