undefined-NaN wrote:

if PQ || MN
>
intersection of MD and BN = K
QM/MK = PN/NK

F a point in BC line and KF 90DEGREE with BC
E a point in DC line and KE 90DEGREE with DC
QB/BF=QM/MK SO MB/KF = QB/QB+BF =QM/QM+MK
PN/NK=PD/DE SO DN/KE= PN/NK+PN = QM/MK+QM
also MB=DN so KF=KE so KFCE = a square so CK is a half angle line of 90 degree
so BK/KP = BC/CP
so QK/KD = QC/DC

I will continue after from here...

ı SOLVED

let draw a line for BD
BD must be parellel to PQ || MN
because AB || DC
CB/CQ = BD/PQ
also CD/CP=AB/CP
SO ABN=~CPB
T, intersection of AB PQ
so BP/BN = BM/BT
so you can see MN paralel to PQ

also we had another way here
BM // DC so CB/CQ = AD/CQ
also CD/CP = AB/CP
so here ABD=~CPQ
so hipo of ABD = BD
so hipo of CPQ = PQ
so BD//PQ

also I can check it with half angle here
if NM || PQ
m(QPB)=m(PBD)
m(PQD)=m(QDB)
so BD / QP= KP/KB = KQ/KD
also I was found KC is a half angle of 90degree
so KB/KP = BC/CP