根据附件图片内容，解释”几何定值问题“？---**第21讲 几何定值问题** 在平面几何中有一类问题涉及线段之间的和、差、积、商为定值，涉及角与角之间的数量不变性，涉及几何图形的面积不变性等. 这类问题叫几何定值问题. **【例1】** 如图21-1,设C为定圆上定弧$\widehat{AB}$的中点,$P$为$\widehat{AB}$上任意一点,且$C$与$P$不在直线$AB$的同侧,求证: $\frac{PA + PB}{PC}$为定值. **分析** 先探明定值是什么. 由点$P$的任意性,考虑点$P$与点$A$重合,则$\frac{PA + PB}{PC} = \frac{0 + AB}{AC} = \frac{AB}{AC}$. 因为$A$、$B$、$C$为定点,则$AB$、$AC$为定值,所以问题中定值可能是$\frac{AB}{AC}$. 尝试证明 $\frac{PA + PB}{PC} = \frac{AB}{AC}$, 转化$PA + PB$,寻找相似三角形. **证明** 如图21-2, 在$AP$的延长线上取点$E$,使$PE = PB$, 连接$BE$、$BC$. 因为$C$为$\widehat{AB}$的中点, 所以$\angle APB = 2\angle CPB$. 而$\angle APB = \angle PBE + \angle PEB$, 于是有$\angle CPB = \angle PEB$, 又$\angle PAB = \angle PCB$, 从而$\triangle ABE \backsim \triangle CBP$, 得$\frac{AE}{CB} = \frac{AB}{PC}$, 即$\frac{PA + PE}{CB} = \frac{AB}{PC}$, 因为$PE=PB$, 所以$\frac{PA + PB}{CB} = \frac{AB}{PC}$, 则$\frac{PA + PB}{PC} = \frac{AB}{CB}$. 因为$C$为$\widehat{AB}$的中点, 所以$CB = AC$. 故$\frac{PA + PB}{PC} = \frac{AB}{AC} = $ 定值. **说明** (1) 上述证法是较为普遍的证法. 对于有关线段比例式(积式)的定值问题,恰当地运用三角形相似是有益的. (2) 其实,还可以利用托勒密定理解决这个问题. 因为$C$为$\widehat{AB}$的中点,所以$BC = AC$. 对于圆内接四边形$APBC$, 有$AP \times BC + PB \times AC = PC \times AB$, 自然地有$\frac{AP \times BC + PB \times AC}{PC} = AB$. 因为$BC = AC$, 则$\frac{AP \times AC + PB \times AC}{PC} = AB$, $\frac{(AP+PB)AC}{PC} = AB$, 所以$\frac{AP+PB}{PC} = \frac{AB}{AC}$. 由例1 可以知道,解决几何定值问题,常考虑特殊图形数量的不变性得到定值(定量、定位); 或者运用运动变化观点考察运动变化至极限位置时的定值性,得到一般问题的定值. 一旦得知具体定值,就将定值问题转化为证明问题,此时运用证明题的基本思路就可以容易地解决. --- **Chart Description:** **Figure 21-1:** * Type: Geometric figure (circle with inscribed triangle and points). * Elements: * A circle. * Points A, B, C, and P on the circle. * Arc $\widehat{AB}$ is part of the circle. * C is labeled as the midpoint of arc $\widehat{AB}$. * P is a point on arc $\widehat{AB}$. C and P are on opposite sides of line segment AB. * Line segments are drawn connecting A to B, A to C, B to C, A to P, and B to P, and C to P. * Labels: Points A, B, C, P. Figure label "图 21-1". **Figure 21-2:** * Type: Geometric figure (circle with inscribed triangle and external point). * Elements: * A circle. * Points A, B, C, and P on the circle. * Point E is outside the circle, on the extension of line segment AP. * Line segments A to B, A to P, P to E form a line segment AE. * Line segments are drawn connecting A to B, A to C, B to C, A to P, B to P, C to P. * Dashed line segments are drawn connecting B to E and B to C. P is connected to E. * Labels: Points A, B, C, P, E. Figure label "图 21-2".