Question

1. examine the given figure.

$overline{bc}$ is a diameter of circle $o$ and $overline{ab} \cong \overline{ao}$. $\triangle abc$ is inscribed in the circle.
which of the following is a valid conjecture based on the given information?

options:

• $\triangle aco$ is an acute triangle whose angles measure $90^\circ$-$60^\circ$-$30^\circ$
• $\triangle aco$ is an isosceles triangle whose base angles measure $30^\circ$
• the measure of angle $abc$ is $90^\circ$
• $m\overarc{oc} = \frac{1}{2} m\overarc{ac}$

Explanation:

Step1: Analyze given info

Given \( BC \) is diameter, so \( OA = OC \) (radii), \( AB \cong AO \), so \( AB = AO = OC \). \( \triangle ABC \) is inscribed, so \( \angle BAC = 90^\circ \) (Thales' theorem).

Step2: Evaluate options

• Option 1: \( \triangle ACO \) has \( OA = OC \), so isosceles, not acute with \( 90^\circ - 60^\circ - 30^\circ \) (invalid).
• Option 2: \( \triangle ACO \): \( OA = OC \), so isosceles. Let \( AB = AO = r \), \( OB = OA = r \), so \( AB = OB = OA \), \( \triangle ABO \) equilateral, \( \angle AOB = 60^\circ \), \( \angle AOC = 120^\circ \), base angles \( \frac{180 - 120}{2} = 30^\circ \) (valid).
• Option 3: \( \angle ABC \): \( \triangle ABO \) equilateral, \( \angle ABO = 60^\circ \), not \( 90^\circ \) (invalid).
• Option 4: \( m\widehat{OC} \) is 0 (radius, not arc), \( m\widehat{AC} \) related to \( \angle AOC = 120^\circ \), so invalid.

Answer:

\( \triangle ACO \) is an isosceles triangle whose base angles measure \( 30^\circ \)