Beth is planning a playground and has decided to place the swings in such a way that they are the same distance from the jungle gym and the monkey bars. If Beth places the swings at point D, how could she prove that point D is equidistant from the jungle gym and monkey bars?

If segment AD ≅ segment CD, then point D is equidistant from points A and B because congruent parts of congruent triangles are congruent.

If segment AD ≅ segment CD, then point D is equidistant from points A and B because a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects.

If m∠ACD = 90° then point D is equidistant from points A and B because congruent parts of congruent triangles are congruent.

If m∠ACD = 90° then point D is equidistant from points A and B because a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects.

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions

kbmflmklbhml; bmlzb; bdhzz

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)