Question 682141
 Solution :

In triangle ABC, Altitude from B and C are equal with each other. hence we have:
BQ = PC 

now consider two triangle PBC and Triangle QCB 
 we have <BPC = <CQB = 90 degree each, 
BC = CB    is the common sides. 
BQ = PC     given equal altitudes.

Hence by RHS rule , 
Triangle PBC is congruent to Triangle QCB.
PB = QC  by ( CPCT )

also these two triangles are congruent then there height must be equal , hence we 
can say that they must lie between the parallel lines ( PQ and BC ) 
hence PBQC is the ISOSCELES Trapezoid  ( since PQ parallel BC and PB = QC  )