26. (本小题满分10分)
如图①,已知在平行四边形$ABCD$中,$AB = 5$,$BC = 8$,$\cos B= \frac{4}{5}$,$AC$为对角线,$AH\perp BC于点H$,$P是边BC$上的动点,以$CP为半径的\odot C与边AD交于点E$,$F$(点$F在点E$的右侧),射线$CE与射线BA交于点G$.

(1) $AH= $______,$AC= $______.
(2) 当$\angle AGE= \angle AEG$时,求$\odot C$的半径.
(3) 如图②,连接$AP$,当$AP// CG$时,求弦$EF$的长.

(1)3,5
(2)过点E作$EN\perp BC$于点N,如图①. $\because$四边形ABCD为平行四边形,$\therefore AD// BC$,$\therefore \angle AEG=\angle BCG$.$\because \angle AGE=\angle AEG$,$\therefore \angle BCG=\angle AGE$,$\therefore BG=BC=8$,$\therefore AG=BG-AB=8-5=3$,$\therefore AE=AG=3$.易得四边形AHNE为矩形,$\therefore HN=AE=3$,$EN=AH=3$.$\because BH=AB\cdot\cos B=4$,$\therefore CH=BC-BH=8-4=4$,$\therefore CN=CH-HN=4-3=1$.在$Rt\triangle CEN$中,$CE=\sqrt{CN^{2}+EN^{2}}=\sqrt{1^{2}+3^{2}}=\sqrt{10}$.所以$\odot C$的半径为$\sqrt{10}$.
(3)如图②,连接EP交AC于点M,作$CQ\perp EF$于点Q,则$EQ=FQ$. $\because AP// CE$,$AE// PC$,$\therefore$四边形APCE为平行四边形.又$\because CE=CP$,$\therefore$四边形APCE是菱形,$\therefore AC\perp EP$,$CM=AM$,$\therefore CM=\frac{1}{2}AC=\frac{5}{2}$.由
(1)得$AB=AC$,$\therefore \angle ACB=\angle B$.在$Rt\triangle PCM$中,$\because \cos\angle MCP=\frac{CM}{CP}=\cos B=\frac{4}{5}$,$\therefore CP=\frac{\frac{5}{2}}{\frac{4}{5}}=\frac{25}{8}$,$\therefore CE=\frac{25}{8}$.在$Rt\triangle CEQ$中,$CQ=AH=3$,$\therefore EQ=\sqrt{CE^{2}-CQ^{2}}=\sqrt{(\frac{25}{8})^{2}-3^{2}}=\frac{7}{8}$,$\therefore EF=2EQ=\frac{7}{4}$.