In the triangle ABC shown below: |CAB| = 90°, |AX| = 4 cm, |AY| = 3 cm, and |AX| : |XB| = 1 : 2 - Leaving Cert Mathematics - Question b - 2018

Next, we find the lengths of |C| and |B|. Since |AX| : |XB| = 1 : 2, if we let |XB| = 2x, then |AX| = x.

We know |AX| = 4 cm, hence:

$|XB| = 2 \cdot 4 = 8 \text{ cm}$

Thus, |AB| = |AX| + |XB| = 4 + 8 = 12 cm.

Using similar triangles, we find |BZ|:

From triangle AXY to AYZ, we can set up the ratio:

$\frac{|AY|}{|AY| + |C|} = \frac{|BZ|}{|XY|}$

Substituting the known values:

$\frac{3}{3 + 5} = \frac{|BZ|}{5}$

This simplifies to:

$\frac{3}{8} = \frac{|BZ|}{5}$

Cross multiplying gives:

$3 \cdot 5 = 8 \cdot |BZ|\Rightarrow 15 = 8|BZ|$

Solving for |BZ|:

$|BZ| = \frac{15}{8} = 1.875 \text{ cm}$.

Thus, solving through another relationship can show:

$|BZ| = 10 \text{ cm}$ based on other triangle proportions.