數學~~證明題目~How to prove?

a)AB = AB (common)
ㄥABD = ㄥABC = 90° (given)
ㄥDAB = ㄥCAB (given)△ADB ≌ △ACB (A.S.A.)
b)i) By the result of a) , △PTK ≌ △PTR.So KT = TR ,
LT // QR ,
By mid-pt. theorem , KL = LQ.ii)By the result of a) , △PTK ≌ △PTR.So
PK = PR
PQ + QL + LK = PR
(1/3)PR + QL + LK = PR
QL + LK = (2/3)PRQL = LK (proved) , soQL = (1/3)PR
QL = PQ i.e. Q is the mid point of PL , and
QS // LT ,
by mid-pt. theorem , PS = ST.

a)因為BD是CB的延線且AB垂直於CD,
所以角ABD=90度:
角ABD=角ABC=90度
AB=AB(公共邊)
角DAB=角CAB(已知)
所以三角形ADB於三角形ACB (A.S.A)
b)i)因為根據a)部的答案,
所以三角形PTK全等於三角形PTR
所以KT=TR(全等三角形性質)
因為QR平行於LT(已知)
所以KL=LQ(截線定理)
ii)因為PQ=三分之一PR,
所以PQ=三分之一PK(全等三角形性質)
因為KL=LQ且KL+LQ+PQ=PK,
所以KL=LQ=PQ
又因為LQ=PQ且QR平行於LT
所以PS=ST(截線定理)
以上,希望幫到你! ^.^