Question

find the value of z.

Explanation:

Step1: Check similarity of triangles

Since \(\angle BAC=\angle D\) (given angle - angle similarity condition, as \(\angle B\) is common to \(\triangle ABC\) and \(\triangle DBA\)), \(\triangle ABC\sim\triangle DBA\) by AA (angle - angle) similarity criterion.

Step2: Use similarity ratio

For similar triangles \(\triangle ABC\) and \(\triangle DBA\), the ratio of corresponding sides is equal. That is \(\frac{AB}{DB}=\frac{BC}{AB}\).
Substitute \(AB = 16\), \(DB=12 + 15=27\), \(BC = 12\) (This approach is wrong. Let's use another property.

We know that if in a right - angled triangle (by Pythagoras \(16^{2}+12^{2}=20^{2}\), \(15^{2}+20^{2}=25^{2}\), no. Wait, use the property of similar triangles.

Since \(\triangle ABC\sim\triangle DBA\) (AA similarity, \(\angle BAC=\angle D\) and \(\angle B=\angle B\)), we have \(\frac{AC}{AD}=\frac{AB}{BD}\)

First, find \(BD=12 + 15=27\)

Let \(AC = z\), \(AD\) can be found using Pythagoras in \(\triangle ABD\) (if \(\angle BAD = 90^{\circ}\), no. Wait, use the property of similar triangles \(\triangle ABC\sim\triangle DBA\)

\(\frac{AB}{BD}=\frac{BC}{AB}=\frac{AC}{AD}\)

We know \(AB = 16\), \(BC = 12\), \(BD=27\)

By \(\frac{AB}{BD}=\frac{BC}{AB}\), cross - multiply: \(AB^{2}=BC\times BD\) (not relevant here).

Another way: Since \(\triangle ABC\sim\triangle DBA\), \(\frac{AC}{AD}=\frac{AB}{BD}\)

Let \(AD=x\), \(AC = z\), \(CD = 15\), \(BC = 12\), \(AB = 16\), \(BD=27\)

Also, in right - angled triangles (if \(\angle BAC+\angle CAD = 90^{\circ}\) and \(\angle D+\angle CAD=90^{\circ}\), so \(\angle BAC=\angle D\)), \(\triangle ABC\sim\triangle DBA\)

\(\frac{z}{z + 15}=\frac{16}{27}\)

Cross - multiply: \(27z=16(z + 15)\)

Step3: Solve the equation

$$
LATEXBLOCK0
$$

Wait, no. Wait, correct property: If \(\triangle ABC\sim\triangle DBA\) (AA similarity), then \(\frac{AC}{AB}=\frac{AB}{BD}\)

\(AC\times BD=AB^{2}\)

Substitute \(AB = 16\), \(BD=27\)

\(z\times27=16^{2}\)

\(27z = 256\)

\(z=\frac{256}{27}\approx9.48\)

No, another approach: Use the geometric mean theorem (right - triangle altitude theorem). But the figure is not clearly marked as right - angled. Wait, assume \(\angle BAC+\angle CAD = 90^{\circ}\) and \(\angle D+\angle CAD = 90^{\circ}\), so \(\angle BAC=\angle D\) and \(\angle B\) is common.

\(\triangle ABC\sim\triangle DBA\)

\(\frac{AC}{AB}=\frac{AB}{BD}\)

\(AC=\frac{AB^{2}}{BD}\)

Since \(AB = 16\), \(BD=12 + 15=27\)

\(z=\frac{16^{2}}{27}=\frac{256}{27}\approx9.48\)

Answer:

\(z=\frac{256}{27}\)