maths

Angle AKB = 90 degree (property of rhombus).
So AK//QD.
AK = KD (property of rhombus)
So by mid-point theorem, QD = 2PK = AK since P is the mid-point of AK.
For triangle QKD and triangle DKC
QD = AK = KC (proved)
KD = KD (common)
Angle QDK = angle DKC = 90 degree.
So triangle QKD congruent triangle DKC (SAS)
so QK = DC
so CDQK is a parallelogram ( opposite sides equal).


2012-01-22 13:03:44 補充：
Correction : 3rd line should be BK = KD.

(just concentrate what you don't understand)

BK = KD (properties of rhombus)
triangle BQD is similar to triangle BPK (AAA)
BD/BK = QD/PK => QD/PK = 2 => QD = 2PK
since P is the mid-point of AK => AK = 2PK
thus, AK = QD