in the diagram below, $\overline{ab} \parallel \overline{cd}$ and $\overline{bo} \cong \overline{co}$. which reason justifies that $\triangle aob \cong \triangle doc$?
options:
angle-side-angle
side-angle-side
hypotenuse-leg
side-side-side

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Accepted Answer

# Brief Explanations:
1. Given $ \overline{AB} \parallel \overline{CD} $, so alternate interior angles $ \angle ABO \cong \angle DCO $ (or $ \angle BAO \cong \angle CDO $) by the Alternate Interior Angles Theorem.
2. Given $ \overline{BO} \cong \overline{CO} $.
3. Vertical angles $ \angle AOB \cong \angle DOC $ (vertical angles are congruent).

Now, let's check the congruence criteria:
- Angle - Side - Angle (ASA): We have two angles and the included side. Here, we have $ \angle ABO \cong \angle DCO $, $ \overline{BO} \cong \overline{CO} $, and $ \angle AOB \cong \angle DOC $. Wait, no, actually, if we take $ \angle BAO \cong \angle CDO $ (from parallel lines), $ \overline{AB} $ and $ \overline{CD} $ (but we know $ \overline{BO} \cong \overline{CO} $ and vertical angles. Wait, maybe Side - Angle - Side (SAS)? Wait, no, let's re - evaluate.

Wait, the correct approach:
Since $ \overline{AB} \parallel \overline{CD} $, $ \angle A=\angle D $ (alternate interior angles) and $ \angle B=\angle C $ (alternate interior angles). Also, $ \overline{BO} \cong \overline{CO} $. And $ \angle AOB=\angle DOC $ (vertical angles). But the key is the congruence criterion. Wait, the options are Angle - Side - Angle (ASA), Side - Angle - Side (SAS), Hypotenuse - Leg (HL, which is for right triangles), and Side - Side - Side (SSS).

Wait, let's list the congruent parts:
- $ \angle AOB \cong \angle DOC $ (vertical angles)
- $ \overline{BO} \cong \overline{CO} $ (given)
- $ \angle OBA \cong \angle OCD $ (alternate interior angles, since $ \overline{AB} \parallel \overline{CD} $)

So we have two angles and the included side (the side between the two angles). The side between $ \angle OBA $ and $ \angle AOB $ is $ \overline{BO} $, and the side between $ \angle OCD $ and $ \angle DOC $ is $ \overline{CO} $, and $ \overline{BO} \cong \overline{CO} $. So this is Angle - Side - Angle (ASA)? Wait, no, the Angle - Side - Angle (ASA) has the side included between the two angles. Wait, in $ \triangle AOB $ and $ \triangle DOC $:

- $ \angle AOB=\angle DOC $ (vertical angles)
- $ \overline{BO}=\overline{CO} $ (given)
- $ \angle OBA=\angle OCD $ (alternate interior angles)

So the order is Angle ( $ \angle OBA $) - Side ( $ \overline{BO} $) - Angle ( $ \angle AOB $)? No, that's Angle - Side - Angle? Wait, no, the angle at $ O $ ( $ \angle AOB $) and angle at $ B $ ( $ \angle OBA $) with the included side $ BO $. And in $ \triangle DOC $, angle at $ O $ ( $ \angle DOC $) and angle at $ C $ ( $ \angle OCD $) with included side $ CO $. Since $ \overline{BO} \cong \overline{CO} $, $ \angle AOB \cong \angle DOC $, and $ \angle OBA \cong \angle OCD $, this is Angle - Side - Angle (ASA). Wait, but let's check the options. The first option is Angle - Side - Angle.

Wait, maybe I made a mistake. Let's think again. The correct congruence criterion here is Angle - Side - Angle (ASA). Because we have two angles and the included side. The two angles are $ \angle AOB $ and $ \angle OBA $ (in $ \triangle AOB $) and $ \angle DOC $ and $ \angle OCD $ (in $ \triangle DOC $), with the included side $ BO $ and $ CO $ respectively, and $ BO\cong CO $.

# Answer:
A. Angle - Side - Angle

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