< xcy β
< xby
step-by-step explanation:
in Ξ΄cax and Ξ΄bax,
ac β
ab (given)< cax β
< bax (given)ax β
ax (common side)
so, Ξ΄cax β
Ξ΄bax (side-angle-side or sas)
since Ξ΄cax β
Ξ΄bax, we can conclude < acx β
< abx
in Ξ΄acy and Ξ΄aby,
ac β
ab (given)< cay β
< bay (since < cax β
< bax is given and xy is the extension of line ax)ay β
ay (common side)
so, Ξ΄acy β
Ξ΄aby (side-angle-side or sas)
since Ξ΄acy β
Ξ΄aby, we can conclude < acy β
< aby
now,
< acy β
< aby
=> < acx + < xcy β
< abx + < xby
=> < acx + < xcy β
< acx + < xby (since < acx β
< abx already proved above)
subtracting < acx from both sides, we get
< acx + < xcy -< acx β
< acx + < xby -< acx
cancelling out < acx and -< acx from both sides, we get
< xcy β
< xby (proved)
![Use a paragraph, flow chart, or two-column proof to prove the angle congruency. given: β cxy β
β bxy](/tpl/images/02/06/XoqgO1RH9a0ktYLf.jpg)