[幾何]數學題....20分呀!!!

14.aBC=DA(given)
CD=AB(giveb)
BD=DB(common side)

∴△BCD is congruent to △DAB

∴angle ADB=angle CBD &angle ABD=angleCDB(corr. angles,congruent△s)

b.BC=DA(given)
angle CBM=angle ADM(proved)
angle CMB=angle AMD(vert. opp. angles)

∴△CBM is congruent to △ADM

∴AM=MD &BM=MD(corr. sides.congruent△s)

2006-11-20 21:51:07 補充：
15a.AX=BX(given)OX=OX(common side)∵OX⊥AB∴angle AXO=angleBXO=90∴△AXO is congruent to △BXOb.AO=OB(corr. sides,congruent△s)by similar method we can proved that △AYO is congruent to △CYO∴AO=OC(corr. sides,congruent△s)∴OB=OC

2006-11-20 22:08:47 補充：
16a.CY=AY-AC=AX-AB=BXBC=CB(common side)∵ AB=AC∴angle ACB=angle ABC(base angles,isos.△)angleYCB+angleACB=180(adj. angles on st. line)Angle YCB=180-angle ABC=angle XBC∴△BCY is congruent to △CBX

2006-11-20 22:09:47 補充：
b.∵angle ZCB=andle ZBC(corr. angles,congruent△s)△CZB is isosceles(sides opp. eq. angles)17RA=PD(given)angle RAC=angle PDB(given)AC=AB+BC=BC+CD=DB∴△RACis congruent to △PDB(SAS)∴angle QCB=angleQBC(corr. angle,congruent △)∴BQ=CQ(sides opp. eq. angles)

14)
a)
AD = BC 及 AB = DC (題設)
所以 ABCD 是一平行四邊形 (對邊相等)
∠ADB = ∠CBD (平行線的內錯角)
∠ABD = ∠CDB (平行線的內錯角)
b)
∠ABD = ∠CDB (平行線的內錯角)
∠AMB = ∠CMD (對頂角)
AB = CD (題設)
△AMB ≌ △CMD (AAS)
AM = MC 及 BM = MD (全等三角形的對應邊)
所以AC及BD互相平分

15)
a)
OX = OX (共用)
AX = XB (題設)
∠OXA = ∠OXB (直角，OX⊥AB)
△OAX ≌ △OBX (SAS)
b)
OY = OY (共用)
AY = YB (題設)
∠OYA = ∠OYB (直角，OY⊥AC)
△OAY ≌ △OBY (SAS)
OA = OC(對應邊△OAY ≌ △OBY)
OB = OA (對應邊△OAX ≌ △OBX)
所以OB=OC

16)
△BAC是等腰三角形(AB=AC)
∠ABC=∠ACB (等腰三角形的底角)
∠XBC=∠YCB (等角的補角)
AY – AC = AX – AB (等量減等量)
YC = XB (等量減等量)
CB = BC (共用)
△BCY ≌ △CBX (SAS)
b)
∠YBC=∠XCB (對應角，△BCY ≌ △CBX)
△BCZ是等腰三角形 (底角相等∠YBC=∠XCB)

17)
AB + BC = DC + CB (等量加等量)
AC = DB (等量加等量)
AR = DP (題設)
∠RAD=∠PDA (題設)
△RAC≌△PBD (ASA)
∠QCB=∠QBC (對應角，△RAC ≌ △D PBD)
△BQC為等腰三角形 (底角相等∠QCB=∠QBC)
所以
BQ = CQ (等腰三角形的腰)