《2025年北大绿卡九年级数学下册人教版》

16. 如图，点P是⊙O的直径AB延长线上一点，且AB = 4，点M为$\overset{\frown}{AB}$上一个动点(不与A，B重合)，射线PM与⊙O交于点N(不与M重合)．
(1)当点M在什么位置时，△MAB的面积最大？并求出这个最大值.
(2)求证：△PAN∽△PMB.

答案：
(1)【解】通过分析可知当点$M$在$\overset{\frown}{AB}$的中点处时，
$\triangle MAB$的面积最大，连接$OM$，此时$OM\perp AB$.
$\because OM=\frac{1}{2}AB=\frac{1}{2}\times4 = 2$，
$\therefore S_{\triangle ABM}=\frac{1}{2}AB\cdot OM=\frac{1}{2}\times4\times2 = 4$.
(2)【证明】$\because\angle PAN=\angle PMB,\angle P=\angle P$，
$\therefore\triangle PAN\sim\triangle PMB$.

17. 如图，在四边形ABCD中，AB = AD，AC与BD交于点E，∠ADB = ∠ACB.
(1)求证：$\frac{AB}{AE}=\frac{AC}{AD}$；
(2)若AB⊥AC，AE：EC = 1：2，F是BC中点，求证：四边形ABFD是菱形.

答案： 【证明】
(1)$\because AD = AB$，$\therefore\angle ABD=\angle ADB$.
$\because\angle ADB=\angle ACB$，$\therefore\angle ABD=\angle ACB$.
又$\angle BAE=\angle CAB$，$\therefore\triangle ABE\sim\triangle ACB$，
$\therefore\frac{AB}{AE}=\frac{AC}{AB}$.$\because AD = AB$，$\therefore\frac{AB}{AE}=\frac{AC}{AD}$.
(2)$\because AE:EC = 1:2$，
$\therefore$设$AE = x$，则$EC = 2x$，$AC = 3x$.
由
(1)知$AB^{2}=AE\cdot AC$，$\therefore AB=\sqrt{3}x$.
又$\because AB\perp AC$，$\therefore BC = 2\sqrt{3}x$，$\therefore\angle ACB = 30^{\circ}$.
$\because F$是$BC$的中点，$\therefore BF=\sqrt{3}x$，
$\therefore BF = AB = AD$.
又$\because\angle ADB=\angle ACB=\angle ABD = 30^{\circ},\angle BAC = 90^{\circ}$，
$\therefore\angle ADB=\angle CBD = 30^{\circ}$，$\therefore AD// BF$，
$\therefore$四边形$ABFD$是平行四边形.
又$\because AB = AD$，$\therefore$四边形$ABFD$是菱形.