After proving ADB is congruent to CDB , how can you prove that < ADB is congruent to < CDB ? Please help! I’ll give Brainliest!

pls ask

the answer would be d.

step-by-step explanation:

move all variables to the left side and all constants to the right side

(x−3)^2+(y−1)^2=16

this is the form of a circle. use this form to determine the center and radius of the circle.

( x − h ) 2 + ( y − k ) 2 = r 2

match the values in this circle to those of the standard form. the variable   r   represents the radius of the circle,   h   represents the x-offset from the origin, and   k   represents the y-offset from origin.

r = 4

h = 3

k = 1

the center of the circle is found at   ( h , k ) .

center:   ( 3 , 1 )

these values represent the important values for graphing and analyzing a circle.

center:   ( 3 , 1 ).

radius:   4

is it like a multiple choice? if so then i need the multiple choice so i can get and understanding on what the answers is so that i can explain it to you