Question 6B [AB] and [CD] are chords of a circle that intersect externally at E, as shown - Leaving Cert Mathematics - Question 6B - 2014

To prove this, we can apply the properties of similar triangles.

Given that △ABE ~ △CDE, we can state that the ratios of the corresponding sides are equal:

$\frac{|EA|}{|EC|} = \frac{|EB|}{|ED|}$

Cross-multiplying yields:

$|EA| \cdot |ED| = |EB| \cdot |EC|$

Thus, we can rearrange this to see that:

$|EA| \cdot |EB| = |EC| \cdot |ED|$

This completes the proof.

First, we express |AD| in terms of the other lengths. From the product we previously established:

$|EA| \cdot |EB| = |EC| \cdot |ED|$

Given |EB| = 6.25 and |ED| = 5.94, we can find |EC| by noting that |CB| = 10 implies:

$|EC| = |CB| - |AD| = 10 - |AD|$

Substituting into the equation gives:

$|EA| \cdot 6.25 = (10 - |AD|) imes 5.94$

Since we do not yet know |EA|, we cannot directly solve for |AD| without additional information. Thus, further steps are needed to find a relationship or additional values to isolate |AD|.