anonymous
  • anonymous
Q10: What's wrong in this question?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1340318787218:dw|
anonymous
  • 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
  • maheshmeghwal9
Omg mahmit sir O.O

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anonymous
  • anonymous
Hi , can you solve the problem above?
anonymous
  • anonymous
It shows 120=90 !!
ganeshie8
  • ganeshie8
if AD = AE, PE has to be more than PC. both cannot be same.. so question is wrong.
ganeshie8
  • 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
  • anonymous
PK is perpendicular bisector and shows PE=PC what's wrong?
anonymous
  • anonymous
no P change with E it obviously depend to E
ganeshie8
  • ganeshie8
wait.. im confused again.... gimme some time il get back
anonymous
  • anonymous
ok thanks
ganeshie8
  • ganeshie8
PD = PC (P is on perpendicular bisector of DC)
anonymous
  • anonymous
yes it is
ganeshie8
  • ganeshie8
i see it some what blurred way.. let me explain my logic
ganeshie8
  • ganeshie8
with P as center, and PD as radius draw a circle. it goes through D and C points
ganeshie8
  • ganeshie8
draw another circle, with A as center, and AD as radius
ganeshie8
  • ganeshie8
|dw:1340380572369:dw|
anonymous
  • anonymous
Do you think points P A and E are in a straight line?
ganeshie8
  • ganeshie8
no way lol... PAE & PBC are congruent triangles i can see that
ganeshie8
  • ganeshie8
the thing im trying to figure out and put is : two circles intersect at two no more than two points...
ganeshie8
  • ganeshie8
|dw:1340381146334:dw|
ganeshie8
  • ganeshie8
with respect to PA straight line, if P is fixed, and PE = PD, PE and PD lie on either side of PA
ganeshie8
  • ganeshie8
im not sure.. if this is satisfying for u. but its ok to me to conclude that question is wrong based this....
ganeshie8
  • ganeshie8
|dw:1340383701791:dw|
ganeshie8
  • ganeshie8
so if we move E to the other side of PA straight line, triangles EAP, DAP, CBP are congruent
anonymous
  • anonymous
when you move E should change EC perpendicular bisector then K and P will be different.
ganeshie8
  • 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
  • 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
  • ganeshie8
|dw:1340392907050:dw|
anonymous
  • anonymous
|dw:1340392825081:dw|