1)Let <BAD=a <=角
so, <DCB=a
triangle ADP~= triangle QCB (SAS)
Let <QBC=b
<ADP=b
so,<QBA=180-a-b
<PDB=180-a-b
so PDBQ is parallelogram
2)AB=CD
Let <DAB=<DCB=a
AEB~=DFC (SAS)
let <EBA=b then <FDC=b
Let <ABC=c so <ADC=c
<EBF=c-b=<EDF
3)PO=OR (diag. of//gram)
so PO/2=XO
OR/2=OY
OY=XO
OQ=OS(diag. of//gram)
so XQYS is //gram
4)a)Let <ABC=z
so <ADC=z
<EBC=z-60
<ADF=z-60
AD=BC
BE=FD
EBC~=AFD(SAS)
b)since AFD~=BEC
AF=EC
FC=AE
so AEFC is //gram
發表於 09-12-12 12:42 PM
