Question

if bc bisects ∠dga, find the measure of each numbered angle.

Explanation:

Step1: Find \(\angle 3\)

Since \(BC\) bisects \(\angle DCA\), \(\angle 3=\angle 24^\circ = 24^\circ\) (Definition of angle bisector: it divides an angle into two equal parts).

Step2: Find \(\angle 1\)

In \(\triangle DBC\), the sum of interior angles is \(180^\circ\). We know one angle is \(115^\circ\) and \(\angle 3 = 24^\circ\). So \(\angle 1=180^\circ - 115^\circ - 24^\circ\)

$$
LATEXBLOCK0
$$

Step3: Confirm \(\angle 2\)

In \(\triangle ABC\), the sum of interior angles is \(180^\circ\). We know one angle is \(104^\circ\) and \(\angle 3 = 24^\circ\). So \(\angle 2=180^\circ - 104^\circ - 24^\circ\)

$$
LATEXBLOCK1
$$

(Or since \(BC\) bisects \(\angle DCA\), \(\angle 2=\angle 3 = 24^\circ\) directly from the angle - bisector property)

Answer:

\(\angle 1 = 41^\circ\), \(\angle 2 = 24^\circ\), \(\angle 3 = 24^\circ\)