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

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
suppose 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...!!!!!
 2 years ago

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

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

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
It shows 120=90 !!
 2 years ago

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

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
u 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..
 2 years ago

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

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

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

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
ok thanks
 2 years ago

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

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
yes it is
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
ok 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
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
If 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.
 2 years ago

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

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