The answer is m∠C'B'D' = 95°, CQ = 6 and B'D' = 11
m∠C'B'D = 180° - m∠B'C'D' - m∠B'D'C'
m∠B'C'D = m∠BCD, m∠B'D'C' = m∠BDC (dilation)
m∠C'B'D = 180° - 34° - 51° = 95°
[tex]\frac{BC}{B'C'}=\frac{BD}{B'D'} =\frac{36}{18}=2\\\\B'D'=\frac{1}{2}BD=\frac{1(22)}{2}=11[/tex]
ΔC'P'Q ∼ ΔCPQ
[tex]\frac{C'Q}{CQ}=\frac{C'D'}{CD}=\frac{D'Q}{PQ}\\[/tex]
[tex]CQ=C'Q[/tex] × [tex]\frac{CD}{C'D'}=C'Q[/tex] ×[tex]\frac{BC}{B'C'} =2C'Q=2(3)=6[/tex]
Therefore, m∠C'B'D' = 95°, CQ = 6 and B'D' = 11.
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