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why is cp ¯ in the picture below not a perpendicular bisector?

a. c is equidistant from a and b.

b. ap ¯ is not congruent to pb ¯ .

c. cp ¯ is congruent to pb ¯ .

d. ∠cpb is a right angle.

B

Step-by-step explanation:

AP is not congruent to PB

i want to die

what the hell man, i just want the points lol

< 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)