证明：∵在正方形ABCD中,bp=3pc ,
设pc为k,则bp=3k ,
∵BC=DC ,
所以DC = cp+bp =k+3k =4k .
∵ q 为 DC中点,
∴ dp = pc = 2k
则 qc:cp =ad :dq =2 又∵ ∠ADC=∠PCQ =90°
∴△PCQ∽△ADQ