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