- anonymous

Q10:
What's wrong in this question?

- katieb

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- anonymous

|dw:1340318787218:dw|

- anonymous

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

- maheshmeghwal9

Omg mahmit sir O.O

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## More answers

- anonymous

Hi ,
can you solve the problem above?

- anonymous

It shows 120=90 !!

- ganeshie8

if AD = AE,
PE has to be more than PC. both cannot be same.. so question is wrong.

- ganeshie8

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..

- anonymous

PK is perpendicular bisector and shows PE=PC what's wrong?

- anonymous

no P change with E it obviously depend to E

- ganeshie8

wait.. im confused again.... gimme some time il get back

- anonymous

ok thanks

- ganeshie8

PD = PC (P is on perpendicular bisector of DC)

- anonymous

yes it is

- ganeshie8

i see it some what blurred way.. let me explain my logic

- ganeshie8

with P as center, and PD as radius draw a circle.
it goes through D and C points

- ganeshie8

draw another circle, with A as center, and AD as radius

- ganeshie8

|dw:1340380572369:dw|

- anonymous

Do you think points P A and E are in a straight line?

- ganeshie8

no way lol... PAE & PBC are congruent triangles i can see that

- ganeshie8

the thing im trying to figure out and put is : two circles intersect at two no more than two points...

- ganeshie8

|dw:1340381146334:dw|

- ganeshie8

with respect to PA straight line,
if P is fixed, and PE = PD,
PE and PD lie on either side of PA

- ganeshie8

im not sure.. if this is satisfying for u. but its ok to me to conclude that question is wrong based this....

- ganeshie8

|dw:1340383701791:dw|

- ganeshie8

so if we move E to the other side of PA straight line,
triangles EAP, DAP, CBP are congruent

- anonymous

when you move E should change EC perpendicular bisector then K and P will be different.

- ganeshie8

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

- anonymous

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.

- ganeshie8

|dw:1340392907050:dw|

- anonymous

|dw:1340392825081:dw|