1. (2024 春·新吴区月考)等腰三角形的两条边长分别为 6 cm 和 10 cm,则周长为
26或22
cm.答案：26或22
2. 已知等腰三角形的一个外角等于 $100^{\circ}$,则它的底角等于
$80^{\circ}$或$50^{\circ}$
.答案：$80^{\circ}$或$50^{\circ}$
3. (2024·海陵区期末)已知 $\triangle ABC$ 是等腰三角形,若 $\angle A = 20^{\circ}$,则 $\angle B$ 的度数为
$20^{\circ}$或$80^{\circ}$或$140^{\circ}$
.答案：$20^{\circ}$或$80^{\circ}$或$140^{\circ}$
4. 若等腰三角形一腰上的高与另一腰的夹角为 $48^{\circ}$,则其底角的度数为
$69^{\circ}$或$21^{\circ}$
.答案：$69^{\circ}$或$21^{\circ}$
5. (2024 春·高新区月考)已知等腰三角形的周长为 15 cm,一腰上的中线把等腰三角形分成周长之差为 3 cm 的两个三角形,则腰长为
4cm或6cm
.答案：4cm或6cm
6. 已知 $\triangle ABC$ 是等腰三角形,过 $\triangle ABC$ 的一个顶点的一条直线,把 $\triangle ABC$ 分成两个小三角形,如果这两个小三角形也是等腰三角形,试求出 $\triangle ABC$ 各内角的度数. (写出所有的情况)
答案：
解:一共有4种情况.
如答图①,$\triangle ABC$是等腰三角形,$AB = AC$,线段$AD$过顶点$A$.根据题意,知$\triangle ABD,\triangle ACD$是等腰三角形,且$AD = BD$,$AD = CD$,那么$\angle B=\angle BAD=\angle CAD=\angle C$,利用三角形内角和定理,可知$\angle B+\angle BAD+\angle CAD+\angle C = 180^{\circ}$,解得$\angle B=\angle BAD=\angle CAD=\angle C = 45^{\circ}$,$\angle BAC = 90^{\circ}$.
如答图②,$AB = AC = CD$,$AD = BD$,设$\angle B = x$,则$\angle BAD=\angle ACB = x$,$\therefore\angle ADC = 2x$,$\because AC = CD$,$\therefore\angle CAD = 2x$,$\therefore\angle BAC = 3x$,$\therefore x + 3x + x = 180^{\circ}$,解得$x = 36^{\circ}$,$\therefore\angle B=\angle C = 36^{\circ}$,$\angle BAC = 108^{\circ}$.
如答图③,$AB = AC$,$AD = BD = BC$,设$\angle A = x$,则$\angle ABD = x$,$\therefore\angle BDC = 2x$,$\because BD = BC$,$AB = AC$,$\therefore\angle ABC=\angle ACB = 2x$,$\therefore x + 2x + 2x = 180^{\circ}$,$\therefore x = 36^{\circ}$,$\therefore\angle A = 36^{\circ}$,$\angle ABC=\angle C = 72^{\circ}$.
如答图④,$AB = AC$,$BC = CD$,$AD = BD$.设$\angle A = x$,则$\angle ABD = x$,$\angle BDC=\angle DBC = 2x$,$\therefore\angle ABC=\angle C = 3x$,$\therefore x + 3x + 3x = 180^{\circ}$,$\therefore x=\frac{180^{\circ}}{7}$,$\therefore\angle A=\frac{180^{\circ}}{7}$,$\angle ABC=\angle C=\frac{540^{\circ}}{7}$.
解:一共有4种情况.
如答图①,$\triangle ABC$是等腰三角形,$AB = AC$,线段$AD$过顶点$A$.根据题意,知$\triangle ABD,\triangle ACD$是等腰三角形,且$AD = BD$,$AD = CD$,那么$\angle B=\angle BAD=\angle CAD=\angle C$,利用三角形内角和定理,可知$\angle B+\angle BAD+\angle CAD+\angle C = 180^{\circ}$,解得$\angle B=\angle BAD=\angle CAD=\angle C = 45^{\circ}$,$\angle BAC = 90^{\circ}$.
如答图②,$AB = AC = CD$,$AD = BD$,设$\angle B = x$,则$\angle BAD=\angle ACB = x$,$\therefore\angle ADC = 2x$,$\because AC = CD$,$\therefore\angle CAD = 2x$,$\therefore\angle BAC = 3x$,$\therefore x + 3x + x = 180^{\circ}$,解得$x = 36^{\circ}$,$\therefore\angle B=\angle C = 36^{\circ}$,$\angle BAC = 108^{\circ}$.
如答图③,$AB = AC$,$AD = BD = BC$,设$\angle A = x$,则$\angle ABD = x$,$\therefore\angle BDC = 2x$,$\because BD = BC$,$AB = AC$,$\therefore\angle ABC=\angle ACB = 2x$,$\therefore x + 2x + 2x = 180^{\circ}$,$\therefore x = 36^{\circ}$,$\therefore\angle A = 36^{\circ}$,$\angle ABC=\angle C = 72^{\circ}$.
如答图④,$AB = AC$,$BC = CD$,$AD = BD$.设$\angle A = x$,则$\angle ABD = x$,$\angle BDC=\angle DBC = 2x$,$\therefore\angle ABC=\angle C = 3x$,$\therefore x + 3x + 3x = 180^{\circ}$,$\therefore x=\frac{180^{\circ}}{7}$,$\therefore\angle A=\frac{180^{\circ}}{7}$,$\angle ABC=\angle C=\frac{540^{\circ}}{7}$.
7. 已知 $\triangle ABC$ 中, $\angle A = 80^{\circ}$,过 $\triangle ABC$ 的顶点 $B$ 的直线将 $\triangle ABC$ 分割成两个等腰三角形,求 $\angle C$ 的度数. (请画图分析)
答案：
解:如答图①,$\because AB = BD = CD$,$\angle A = 80^{\circ}$,$\therefore\angle ADB=\angle A = 80^{\circ}$,$\angle DBC=\angle C$,$\because\angle ADB=\angle DBC+\angle C$,$\therefore\angle C=\frac{1}{2}\angle ADB = 40^{\circ}$;
如答图②,$\because AB = AD$,$CD = BD$,$\angle A = 80^{\circ}$,$\therefore\angle ADB=\angle ABD=\frac{1}{2}(180^{\circ}-\angle A)=50^{\circ}$,$\angle DBC=\angle C$,$\because\angle ADB=\angle DBC+\angle C$,$\therefore\angle C=\frac{1}{2}\angle ADB = 25^{\circ}$;
如答图③,$\because AD = BD = CD$,$\therefore\angle A=\angle DBA = 80^{\circ}$,$\therefore\angle C=\angle DBC=\frac{180^{\circ}-2\times80^{\circ}}{2}=10^{\circ}$.
综上所述,$\angle C$的度数为$40^{\circ}$或$25^{\circ}$或$10^{\circ}$.
解:如答图①,$\because AB = BD = CD$,$\angle A = 80^{\circ}$,$\therefore\angle ADB=\angle A = 80^{\circ}$,$\angle DBC=\angle C$,$\because\angle ADB=\angle DBC+\angle C$,$\therefore\angle C=\frac{1}{2}\angle ADB = 40^{\circ}$;
如答图②,$\because AB = AD$,$CD = BD$,$\angle A = 80^{\circ}$,$\therefore\angle ADB=\angle ABD=\frac{1}{2}(180^{\circ}-\angle A)=50^{\circ}$,$\angle DBC=\angle C$,$\because\angle ADB=\angle DBC+\angle C$,$\therefore\angle C=\frac{1}{2}\angle ADB = 25^{\circ}$;
如答图③,$\because AD = BD = CD$,$\therefore\angle A=\angle DBA = 80^{\circ}$,$\therefore\angle C=\angle DBC=\frac{180^{\circ}-2\times80^{\circ}}{2}=10^{\circ}$.
综上所述,$\angle C$的度数为$40^{\circ}$或$25^{\circ}$或$10^{\circ}$.
8. 如图,在 $\triangle ABC$ 中, $\angle B = 50^{\circ}$, $\angle C = 90^{\circ}$,在射线 $BA$ 上找一点 $D$,使 $\triangle ACD$ 为等腰三角形,则 $\angle ACD$ 的度数为
$70^{\circ}$或$40^{\circ}$或$20^{\circ}$
.答案：$70^{\circ}$或$40^{\circ}$或$20^{\circ}$
9. 如图,在等腰三角形 $ABC$ 中, $AB = AC$, $\angle B = 50^{\circ}$, $D$ 为 $BC$ 的中点,点 $E$ 在 $AB$ 上, $\angle AED = 70^{\circ}$,若 $P$ 是等腰三角形 $ABC$ 的腰上的一点,则当 $\triangle DEP$ 是以 $\angle EDP$ 为顶角的等腰三角形时, $\angle EDP$ 的度数是
$40^{\circ}$或$100^{\circ}$或$140^{\circ}$
.答案：$40^{\circ}$或$100^{\circ}$或$140^{\circ}$