< xcy ≅ < xby
step-by-step explanation:
in δcax and δbax,
ac ≅ ab (given)< cax ≅ < bax (given)ax ≅ ax (common side)
so, δcax ≅ δbax (side-angle-side or sas)
since δcax ≅ δbax, we can conclude < acx ≅ < abx
in δacy and δaby,
ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)
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)