问题1.在等腰三角形ABC中，AB = AC，∠B和∠C的等分线在O处相交。将A连到O。证明：
(i)OB = OC(ii)AO对分∠A
解决方案：
Given: (i) An isosceles ∆ABC in which AB=AC
(ii) bisects of ∠B and ∠C each other at O.
Show: (i) OB=OC
(ii) AO bisects ∠A (∠1=∠2)
(i) In ∆ABC,
AB = AC
∠B =∠C [angles opposite to equal sides are equal]
1/2 ∠B = 1/2∠C
∠OBC=∠OCB
∴OB = OC [sides opposite equal ∠ are equal]
(ii) In ∆AOB and ADC
AB = AC [given side]
1/2 ∠B = 1/2∠C
∠ABO = ∠ACO [Angle]
BO = OC [proved above side]
∴∆AOB ≅ AOC
Thus ∠1 = ∠2
Therefore, AO bisects ∠A
问题2.在ΔABC中，AD是BC的垂直平分线(见图)。证明ΔABC是一个等腰三角形，其中AB = AC。
解决方案：
Given: AD is ⊥ bisector of BC
Show: AB=BC
In ∆ABD and ∆ACD
BD=DC [AD is ⊥ bisector side]
∠ADB=∠ADC [Each 90° angle]
AD=AD [common side]
∴∆ABD≅∆ACD [S.A.S]
AB=AC [C.P.C.T]
问题3。ABC是一个等腰三角形，其中BE和CF的高度分别画在AC和AB的相等边上(见图)。证明这些高度相等。
解决方案：
Given: AB=AC,BE and CF are altitudes
Show: BE=CF
In ∆AEB and ∆AFC,
∠E=∠F [Each 90° angle]
∠A=∠A [common angle]
AB=AC [given side]
∴∆AEB≅∆AFC [A.A.S]
BE=CF [C.P.C.T]
问题4. ABC是一个三角形，其中AC和AB边的BE和CF高度相等(见图)。显示
(i)ΔABE≅ΔACF
(ii)AB = AC，即ABC是等腰三角形。
解决方案：
Given: Altitudes BE and CF to sides AC and AB are equal
Show: (i) ΔABE ≅ ΔACF
(ii) AB = AC
(i) In ∆ABF and ∆ACF,
∠E=∠F [Each 90° angle]
∠A=∠A [common angle]
AB=AC [given] S
∴∆AEB≅∆AFC [A.A.S]
(ii) AB=AC [C.P.C.T]
问题5. ABC和DBC是在同一基础BC上的两个等腰三角形(见图)。证明∠ABD=∠ACD。
解决方案：
Given: AB=AC,BD=DC
Show: ∠ABD = ∠ACD
In ∆ABD,
AD=AC
∴∠1=∠2 [angle opposite to equal sides are equal] [1]
In ∆BDC,
BD=DC
∴∠3=∠4 [angle opposite to equal sides are equal] [2]
Adding 1 and 2
∠1+∠2= ∠2+∠4
∠ABD=∠ACD
问题6.ΔABC是一个等腰三角形，其中AB = AC。边BA生成为D，使得AD = AB(见图)。证明∠BCD是一个直角。
解决方案：
In ∆ABC,
AB=AC
∠ACB=∠ABC [1]
In ∆ACD,
AC=AD
∠ACD=∠ADC [2]
Adding 1 and 2
∠ACB+∠ACD=∠ABC+∠ADC
∠BCD=∠ABC+∠BDC
Adding ∠BCD on both side
∠BCD+∠BCD=∠ABC+∠BDC+∠BCD
2∠BCD=180°
∠BCD=(180°)/2=90°
问题7. ABC是一个直角三角形，其中∠A= 90°，AB = AC。找到∠B和∠C。
解决方案：
Find: ∠B=? and ∠C?
In ∆ABC,
AB=AC
∴∠B=∠C [angle opposite to equal side are equal]
∠A+∠B+∠C=180° [angle sum property of triangle]
90°+∠B+∠B=180°
2∠B=180°-90°
∠B=(90°)/2=45°
Therefore, ∠B=45° and ∠C =45°
问题8：证明等边三角形的角度均为60°。
解决方案：
Given: Let ∆ABC is an equilateral ∆
Show: ∠A=∠B=∠C=60°
In ∆ABC,
AB=AC
∠B=∠C [1]
Also
AC=BC
∠B=∠A [2]
From 1,2 and 2
∠A=∠B=∠C
In ∆ABC,
∠A+∠B+∠C=180° [angle sum property of triangle]
∠A+∠A+∠A=180°
3∠A=180°
∠A=(180°)/3=60°
∠A=60°
∴∠B=60° and ∠C=60°