Geometry Setup

We are given an isosceles triangle $\Delta ABC$ with base $AB$. The length of the base is $AB = 8$ units.

An altitude $CD$ is drawn from vertex $C$ to the base $AB$, with length $CD = 4$ units. Since $\Delta ABC$ is isosceles with $AC = BC$, the altitude $CD$ bisects the base $AB$. Therefore, point $D$ is the midpoint of $AB$.

This means $AD = DB = \frac{AB}{2} = \frac{8}{2} = 4$ units.

Analyzing Right-Angled Triangle ADC

Consider the triangle $\Delta ADC$. We know the following lengths:

• $AD = 4$ units
• $CD = 4$ units
• $\angle ADC = 90^\circ$ (since $CD$ is the altitude)

Since $AD = CD$, $\Delta ADC$ is an isosceles right-angled triangle.

Calculating Angles

In $\Delta ADC$, the sum of angles is $180^\circ$. We have:

$\angle CAD + \angle ACD + \angle ADC = 180^\circ$

Since $\Delta ADC$ is isosceles with $AD = CD$, the angles opposite these sides are equal:

$\angle CAD = \angle ACD$

Substituting into the sum of angles equation:

$\angle CAD + \angle CAD + 90^\circ = 180^\circ$

$2 \angle CAD = 180^\circ - 90^\circ$

$2 \angle CAD = 90^\circ$

$\angle CAD = 45^\circ$

Therefore, $\angle ACD = 45^\circ$.

Determining Angles of Triangle ABC

The angles of the triangle $\Delta ABC$ are:

• $\angle CAB$: This is the same angle as $\angle CAD$. So, $\angle CAB = 45^\circ$.
• $\angle ABC$: Since $\Delta ABC$ is isosceles with base $AB$, the base angles are equal. $\angle ABC = \angle CAB$. So, $\angle ABC = 45^\circ$.
• $\angle BCA$: This is the vertex angle. The sum of angles in $\Delta ABC$ is $180^\circ$. $\angle BCA = 180^\circ - (\angle CAB + \angle ABC)$ $\angle BCA = 180^\circ - (45^\circ + 45^\circ)$ $\angle BCA = 180^\circ - 90^\circ$ $\angle BCA = 90^\circ$

The set of angles is $\angle CAB, \angle ABC, \angle BCA = 45^\circ, 45^\circ, 90^\circ$.