Question 1210145: In the diagram below, ABCD is a rectangle with AD=6 and B=10. Point M is the intersection of its diagonals and point E lies on \overline{AD} such that BE=2. Find CE^2. (Diagram not to scale)
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let ABCD be a rectangle with AD=6 and AB=10. Let M be the intersection of the diagonals AC and BD. Let E be a point on AD such that BE=2. We want to find CE
2
.
Since ABCD is a rectangle, AD=BC=6 and AB=CD=10.
Since E lies on AD and AD=6, we have 0≤AE≤6.
In right triangle ABE, we have AE
2
+AB
2
=BE
2
.
We are given BE=2 and AB=10, so AE
2
+10
2
=2
2
, which means AE
2
+100=4. This gives AE
2
=−96, which is impossible.
There must be an error in the problem statement.
Let's proceed assuming BE>10.
We have AE
2
+AB
2
=BE
2
.AE
2
+10
2
=BE
2
, so AE
2
+100=BE
2
.AE=
BE
2
−100
.
Since E lies on AD, we have ED=AD−AE=6−AE.
In right triangle CDE, we have CD
2
+ED
2
=CE
2
.CE
2
=10
2
+(6−AE)
2
=100+(6−
BE
2
−100
)
2
.
If BE=2, we have AE
2
=4−100=−96, which is impossible.
If we assume BE=12, then AE
2
=12
2
−10
2
=144−100=44, so AE=
44
=2
11
.ED=6−2
11
.CE
2
=10
2
+(6−2
11
)
2
=100+36−24
11
+44=180−24
11
.
If we assume BE=11, then AE
2
=11
2
−10
2
=121−100=21, so AE=
21
.ED=6−
21
.CE
2
=10
2
+(6−
21
)
2
=100+36−12
21
+21=157−12
21
.
Since the problem states BE=2, and this leads to an impossible situation, we must assume there is an error. However, we can proceed with the methodology.
If we assume AE=x, then BE
2
=x
2
+100.ED=6−x.CE
2
=10
2
+(6−x)
2
=100+36−12x+x
2
=136−12x+x
2
.BE=2, so x
2
+100=4, so x
2
=−96, which is impossible.
Let's assume the diagram has an error and BE is longer.
Let BE=k.
Then AE=
k
2
−100
.CE
2
=10
2
+(6−
k
2
−100
)
2
.
However, the problem states BE=2, so there is an error in the diagram or given lengths.
Final Answer: The final answer is
136
assuming AE=0 which implies BE=10.
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
The problem in the post is written INCORRECTLY.
Indeed, it says " B=10 ".
But 'B' is the point, and writing " B=10 " makes no mathematical sense.
Since the problem is written so inaccurately, we can not accept it for execution.
I am very disappointed to see such inaccurate writing.
In Mathematics, respect for the reader means writing accurately.
Inaccurate writing is treated as disrespect for the reader.
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