Pls help!!
angle pairs!!

< 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)

answer: can you give the

step-by-step explanation:

question says to graph a triangle (xyz) and reflect it over the line y=x to create triangle x'y'z'.

we know that line y=x has both coordinates equal like (1,1), (5,5) etc so we can easily graph that reflecting line.

description of transformation:

we can see that x and y coordinate gets switched after transformation like (2,8) changed to (8,2)

what do you notice about the two line segments you drew:

we see that line segments xp and x'p are of same length and perpendicular to the reflecting line y=x.

do you think you would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line? :

yes, i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line.

how do you know? :

by measuring the length of yk and y'k.