Question

Given circle \(A B C D\) with \(A B \| E F\).

a) Prove \(C D E F\) is a cylindrical quad.
b) If \(\hat{D}_{2}=38^{\circ}\), calculate \(\hat{E}_{2}\)

Answer 1

Solutions:

a) \(\hat{B}_{1}=\hat{C}_{1}\) (_'s same seg.)
\[
\begin{aligned}
&\hat{B}_{1}=\hat{F}_{1} \text { (corres. } \angle \text { 's, AB\|EF) } \\
&\therefore \hat{C}_{1}=\hat{F}_{1}
\end{aligned}
\]
\(\therefore C D E F\) cyc - quad (line subt \(=\angle s\) )

b) \(\hat{D}_{2}=\hat{E}_{2}=38^{*}\) (L's same seg quad CDEF)