AB and CD are two chords in a circle such that AB CD and AB is perpendicular to CD at

I think it is a right angled triangle..how come it cn b similar to APE,ADE etc.as A,D r points on the circle.dey cant form a triangle.plz explain if I am wrong
10 years agoHelpfull: Yes(2) No(3)
ans is : BPE~(similar to APE).
10 years agoHelpfull: Yes(1) No(6)
by the option it can be solved easily....
10 years agoHelpfull: Yes(1) No(1)
similar to pqr
10 years agoHelpfull: Yes(1) No(0)
ans- triangle ADE
according ques- if we draw a fig. & see triangle BPE is right angle, then in circle if join point A and point D , make a tringle ADE, and ADE form of right angle. then, by RSH technique BPE~ similar to ADE
10 years agoHelpfull: Yes(1) No(0)
i too want to know this answer.....plzz can anyone post the answer
10 years agoHelpfull: Yes(0) No(0)
similar to EPF
10 years agoHelpfull: Yes(0) No(1)
question is wrong. AB is perpendicular bisector of CD and the ans is BPE~AFB
9 years agoHelpfull: Yes(0) No(0)
By angle=angle=angle theroem BPE=ADE.....
9 years agoHelpfull: Yes(0) No(0)

AB and CD are two chords in a circle such that AB>CD and AB is perpendicular to CD at point E.P is a point on CD such that BP is extended and meet to the circle at F then triangle BPE is similar to?