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
a) AEC
b) QDP
c) QAB
d) ABC

• TRIANGLE QAB IS SIMILAR TO TRIANGLE PEB.
  ANGLE (E=Q=90) Q = ANGLE IN SEMI CIRCLE.BY CHORD AB
  ANGLE (B=B)
  THEREFORE BY THE PROPERTY OF SIMILAR TRIANGLE.TWO ANGLES ARE EQUAL HENCE TRIANGLES ARE SIMILAR IN NATURE
• 10 years agoHelpfull: Yes(14) No(0)
• AEC...and here chord AB is diagonal also.
  
• 10 years agoHelpfull: Yes(2) No(2)
• sorry..the ans. wil be QAB
• 10 years agoHelpfull: Yes(2) No(0)
• it's QDP according to the properties of similar triangles
  
• 10 years agoHelpfull: Yes(1) No(3)
• qab angle e=q;
  b=b;a=p;
• 10 years agoHelpfull: Yes(1) No(0)
• hey prakar huw u got qdp please explain it
  
• 10 years agoHelpfull: Yes(0) No(0)