a: Xét ΔDAB và ΔCBD có
\(\hat{DAB}=\hat{CBD}\) (GT)
\(\hat{ABD}=\hat{BDC}\) (hai góc so le trong, AB//CD)
Do đó: ΔDAB~ΔCBD
b: ΔDAB~ΔCBD
=>\(\frac{DA}{CB}=\frac{AB}{BD}=\frac{BD}{CD}\)
=>\(\frac36=\frac{6}{CD}=\frac{2.5}{CB}\)
=>\(CD=6\cdot\frac63=6\cdot2=12\left(\operatorname{cm}\right);CB=2,5\cdot\frac63=2,5\cdot2=5\left(\operatorname{cm}\right)\)

