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

• Triangle BPE will be similar to triangle BQA.(using AAA property: Angle B is common in both. angle PEB = angle BQA = 90,since AB will be the diameter. so the 3rd angle will also be same.)
• 10 years agoHelpfull: Yes(8) No(2)
• qab will be similar triangle because they both share the same angle B
• 10 years agoHelpfull: Yes(2) No(1)
• m4maths where is ur maths.
• 10 years agoHelpfull: Yes(2) No(0)
• it will be AEC,
  as angle AEC= angle Bep
• 10 years agoHelpfull: Yes(0) No(1)
• ans is QDP
  AB PARALLEL TO QD
  ANGLE B=ANGLE Q (ALTERNATING ANGLE)
  angles at s are vertically opposite angle hence they are equal
  angle D and angle E are Right angle
• 10 years agoHelpfull: Yes(0) No(2)
• sorry x+3=25 so x=22
• 10 years agoHelpfull: Yes(0) No(1)
• it must b qdp
• 10 years agoHelpfull: Yes(0) No(0)
• ans is QDP
  
• 10 years agoHelpfull: Yes(0) No(0)