BIG HINT: Use the external angle theorem.
If WS = WT and RS = ST = TU, with R-S-T-U, prove that ∠RWS ∠TWU.
My work so far:
Given: WS = WT and RS = ST = TU, with R-S-T-U
Given that R-S-T-U, we know R-S-T, S-T-U, R-S-U, and R-T-U.
∠WST ∠WTS by the Isosceles Triangle Theorem.
m∠WST = m∠WTS by CPCF (Congruent Parts of Congruent Figures)
m∠WSR + m∠WST = 180, and m∠WTU + m∠WTS = 180.
m∠WSR = m∠WTU by algebra
∠WSR ∠WTU, and therefore,
Triangle WSR Triangle WTU by SAS.
And thus, ∠RWS ∠TWU by CPCF.
Am I even close to having a legitimate proof? I'm pretty sure there is something missing since I did not use the given betweenness relation of R-S-T-U.
Any corrections/suggestions, tips/help is greatly appreciated. Thank you very much for your time.
But anyway, if I can use the external angle theorem, I already know
= since triangle WST is isosceles with WS = WT. So, since = , then wouldn't their complements be equal by the theorem that "2 angles that are supplementary to the same angle have equal measures." And thus by, SAS (WS = WT, = , and RS = TU), triangles WSR and WTU are congruent; so and are also congruent by CPCF.