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mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1340318787218:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1suppose to: ABCD is rectangular AE=AD PK and PH are perpendicular bisector EK=KC , DH=HC In conclusion: AP=BP AE=BC PE=PC <PAB=<PBA Then <PAE=<PBC Or <EAB=<CBA=90...!!!!!

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.0Omg mahmit sir O.O

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1Hi , can you solve the problem above?

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1if AD = AE, PE has to be more than PC. both cannot be same.. so question is wrong.

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1u see ? think of a circle with radius AD and center at A.. since P is fixed point, its length changes as it moves along the circle..

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1PK is perpendicular bisector and shows PE=PC what's wrong?

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1no P change with E it obviously depend to E

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1wait.. im confused again.... gimme some time il get back

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1PD = PC (P is on perpendicular bisector of DC)

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1i see it some what blurred way.. let me explain my logic

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1with P as center, and PD as radius draw a circle. it goes through D and C points

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1draw another circle, with A as center, and AD as radius

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1340380572369:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1Do you think points P A and E are in a straight line?

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1no way lol... PAE & PBC are congruent triangles i can see that

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1the thing im trying to figure out and put is : two circles intersect at two no more than two points...

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1340381146334:dw

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1with respect to PA straight line, if P is fixed, and PE = PD, PE and PD lie on either side of PA

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1im not sure.. if this is satisfying for u. but its ok to me to conclude that question is wrong based this....

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1340383701791:dw

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1so if we move E to the other side of PA straight line, triangles EAP, DAP, CBP are congruent

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1when you move E should change EC perpendicular bisector then K and P will be different.

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1ok did i make myself clear why i think PE should come other side of PA ? in your diagram both PE and PD are on same side of PA which is wrong

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1If you have cad to draw exactly the figure please try to figure out what happen in there? Thank you for helping and hope we will get a logical proofing after all.

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1340392907050:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1340392825081:dw