In the diagram below, AE = 4, EB = 3, and DE = 2. Find CE.
![In the diagram below AE 4 EB 3 and DE 2 Find CE class=](https://us-static.z-dn.net/files/d3f/f993eaed05b10f5d73accb784d1fc208.png)
Answer: Value of CE = 6.
Explanation:
Since we have given that
AE=4
BE=3
DE=2
Let CE be x.
As we know that
When two chords intersect each other inside a circle, the products of their segments are equal.
Here, we can see that each chord is cut into two segments at the point of where they intersect.
So,
[tex]EA.EB=EC.ED\\\\4\times 3=x\times 2\\\\12=2x\\\\\frac{12}{2}=x\\\\6=x[/tex]
Hence, value of CE = 6.