Cho tam giác ABC, có góc A bằng 90 độ + góc B, đường cao CH

$$\begin{gathered} \mathrm{Cho} ∆\mathrm{ABCZ},\mathrm{có} \hat{\mathrm{A}} ={90}^{\mathrm{o}}+\hat{\mathrm{B}},\mathrm{đường} \mathrm{cao} \mathrm{CH}.\mathrm{Chứng} \mathrm{minh} \\ \mathrm{a}.\widehat{\mathrm{CBA}}=\widehat{\mathrm{ACH}} \\ \mathrm{b}.{\mathrm{CH}}^2=\mathrm{BH}.\mathrm{AH} \end{gathered}$$