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1. 已知一个等腰三角形的两条边长分别为 3 cm 和 7 cm，则它的底边长是
3
cm.答案
1.3.
2. 如图，在△ABC 中，AB = AC，BD 平分∠ABC. 若∠ADB = 108°，求∠A 的大小.
答案
2.设$∠ DBC = x°$.
$\because$ $BD$平分$∠ ABC$，
$\therefore$ $∠ ABC = 2∠ DBC = (2x)°$.
$\because$ $AB = AC$，
$\therefore$ $∠ C = ∠ ABC = (2x)°$（等边对等角）.
$\because$ $∠ ADB = 108°$，$∠ ADB = ∠ DBC + ∠ C$（三角形的外角等于与它不相邻的两个内角的和），
$\therefore$ $x + 2x = 108$.
$\therefore$ $x = 36$.
$\therefore$ $∠ ABC = ∠ C = 72°$.
$\therefore$ $∠ A = 180° - ∠ ABC - ∠ C = 180° - 72° - 72° = 36°$（三角形的内角和等于$180°$）.
$\because$ $BD$平分$∠ ABC$，
$\therefore$ $∠ ABC = 2∠ DBC = (2x)°$.
$\because$ $AB = AC$，
$\therefore$ $∠ C = ∠ ABC = (2x)°$（等边对等角）.
$\because$ $∠ ADB = 108°$，$∠ ADB = ∠ DBC + ∠ C$（三角形的外角等于与它不相邻的两个内角的和），
$\therefore$ $x + 2x = 108$.
$\therefore$ $x = 36$.
$\therefore$ $∠ ABC = ∠ C = 72°$.
$\therefore$ $∠ A = 180° - ∠ ABC - ∠ C = 180° - 72° - 72° = 36°$（三角形的内角和等于$180°$）.
3. 如图，已知：在△ABC 中，AC = BC，CD 为边 AB 上的中线，E 为边 BC 上一点，EF⊥AB，垂足为 F. 求证：∠ACD = ∠BEF.
答案
3. $\because$ $AC = BC$，$CD$为边$AB$上的中线，
$\therefore$ $∠ ACD = ∠ BCD$，$CD ⊥ AB$（等腰三角形三线合一）.
$\therefore$ $∠ CDB = 90°$.
$\because$ $EF ⊥ AB$，
$\therefore$ $∠ EFB = 90°$.
$\therefore$ $∠ CDB = ∠ EFB$.
$\therefore$ $CD // EF$（同位角相等，两直线平行）.
$\therefore$ $∠ BCD = ∠ BEF$（两直线平行，同位角相等）.
又$\because$ $∠ ACD = ∠ BCD$，
$\therefore$ $∠ ACD = ∠ BEF$.
$\therefore$ $∠ ACD = ∠ BCD$，$CD ⊥ AB$（等腰三角形三线合一）.
$\therefore$ $∠ CDB = 90°$.
$\because$ $EF ⊥ AB$，
$\therefore$ $∠ EFB = 90°$.
$\therefore$ $∠ CDB = ∠ EFB$.
$\therefore$ $CD // EF$（同位角相等，两直线平行）.
$\therefore$ $∠ BCD = ∠ BEF$（两直线平行，同位角相等）.
又$\because$ $∠ ACD = ∠ BCD$，
$\therefore$ $∠ ACD = ∠ BEF$.
