關於數學等腰三角形問題！！！(急)

係... 一定係
而且, 你比埋個直角我.... 更加係

prove....

ao ^2 + op^2 = ap ^2 (畢氏定理)
ao^2 = ap^2 - op^2 ......(1)

bo^2 + op^2 = bp^2 (畢氏定理)
bo^2 = bp^2 - op^2 .....(2)

given ap = bp
ao^2 = ap^2 - op^2 ......(1)
bo^2 = bp^2 - op^2 .....(2)

ao^2 = ap ^2 - op^2 = bp^2 - op^2 = bo^2
ao^2 = bo^2
ao = bo

(當然, 可以一早開左方先都得)

You can use the Congruent Triangle to solve these.

In Triangle PAO,Triangle PBO

since
PO = PO (Common Side)
AP = BP (Given)
Angle PAB = Angle PBA (base angle,isos.Triangle)
Angle POA = Angle POB = 90 Degrees (adj.Angles on st.line)
ie. Angle APO = Angle BPO (angle sum of triangle)

So Therefore,
Triangle PAO is Congruent to Triangle PBO (S.A.S)

So Therefore,
AO = OB (Corr.sides,Congruent Triangle)

2010-06-29 23:04:46 補充：
Wing Kei 既推論係啱,但係你prove吾到咩係平衡.

PO=PO(common)
AP=BP(given)
角AOP=角BOP(OP 垂直 AB)
so, 三角形AOP=三角形BOP(R.H.S)
so,AO=OB(全等三角形對應邊)

OP=PO(公共邊)

AP=BP(已知)

angleAOP+angleBOP=180(直線上的鄰角)
angleAOP.................=90

so,triaangleAOP=triangleBOP(RHS)

so,AO=OB(全等三角形對應邊)

2010-06-22 16:00:42 補充：
To 樓主,你講明(OP 垂直 AB),angleAOP一定係90o.......

2010-06-22 17:10:38 補充：
如用Pyth.theorem.............

AO^2+OP^2=AP^2(Pyth.theorem)
AO^2 =AP^2-OP^2

BO^2+OP^2=PB^2(Pyth.theorem)
BO^2 =PB^2-OP^2

AP=BP(given)

BO^2=AO^2
BO =AO


有冇明白D?

是........因為是等腰三角形........底也要平衡^^