Question

In a trapezium ABCD, AB || CD and AD = BC. If P is the point of intersection of the diagonals AC and BD, prove that PA × PC = PB × PD.

Answer 1

Draw DE ⊥ AB and CF ⊥ AB 

In ∆s DEA and CFB, we have

AD = BC (Given) 

∠DEA = ∠CFB (each = 90°)

 DE = CF (Distance between two parallels) 

∴ ∆DEA ≅ ∆CFB (R.H.S.) 

⇒ ∠DAE = ∠CBF (c.p.c.t.) ...(i) 

Now ∠D + ∠B = ∠ADC + ∠CBA = ∠ADC + ∠CBF 

= ∠ADC + ∠DAE    (From (i)) 

= 180°     (DC || AB, Sum of co-int. ∠s = 180°) 

⇒ Opposite angles of trapezium ABCD are supplementary. 

⇒ ABCD is a cyclic quadrilateral. 

Thus AC and BD are two chords of the circle circumscribring the trapezium such that they intersect at P. 

Hence, PA × PC = PB × PD.