In a circle with centre ‘O’, AB and CD are two chords such that AB > CD and AB is perpendicular bisector of CD at E. P is a point on CD and when BP is extended it meets the circle at Q. For any point P, the triangle BPE is similar to triangle
option
1) AEC
2) QDP
3) QAB
4) ABC

• ans is QAB
  by drawing circle we use the property of circle that angle in semicircle is 90
  in this way bpe has a common angle at b point and angle aqb=bep=90
  hence it is similar.
• 11 years agoHelpfull: Yes(12) No(5)
• Its answer is AEC.
• 11 years agoHelpfull: Yes(8) No(4)
• Answer is AQB.
• 11 years agoHelpfull: Yes(5) No(7)
• ans is AEC.... bcoz AEC is right angled triangle as that of BPE.
• 11 years agoHelpfull: Yes(2) No(5)
• i think it ic AEC..as it is a r8 angle triangle like BPE
• 11 years agoHelpfull: Yes(1) No(0)