Question 1210145
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.