Question

as shown in the diagram below of \\( \triangle abc, \overline{bc} \\) is extended through \\( d, m \angle a = 70 \\), and \\( m \angle acd = 115 \\).
which statement is true?

1. \\( ac > ab \\)
2. \\( ab > bc \\)
3. \\( bc < ac \\)
4. \\( ac < ab \\)

Explanation:

Step1: Find the measure of ∠ACB

Since ∠ACD and ∠ACB are supplementary angles (they form a linear pair), we know that \(m\angle ACB=180 - m\angle ACD\).
Given \(m\angle ACD = 115^{\circ}\), then \(m\angle ACB=180 - 115=65^{\circ}\).

Step2: Find the measure of ∠B

Using the angle - sum property of a triangle (\(m\angle A+m\angle B+m\angle ACB = 180^{\circ}\)), we substitute \(m\angle A = 70^{\circ}\) and \(m\angle ACB=65^{\circ}\).
So, \(m\angle B=180-(70 + 65)=45^{\circ}\).

Step3: Apply the side - angle relationship in a triangle

In \(\triangle ABC\), the side - angle relationship states that the larger angle is opposite the longer side.
Since \(m\angle ACB=65^{\circ}\), \(m\angle A = 70^{\circ}\), and \(m\angle B = 45^{\circ}\), and the side opposite \(\angle A\) is \(BC\), the side opposite \(\angle B\) is \(AC\), and the side opposite \(\angle ACB\) is \(AB\).
We know that if \(m\angle B\lt m\angle ACB\lt m\angle A\), then \(AC\lt AB\lt BC\) (because in a triangle, \(a=\frac{b\sin A}{\sin B}\) (by the sine rule \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)), and when comparing sides \(a,b,c\) opposite angles \(A,B,C\) respectively).

Answer:

1. \(AC\lt AB\)