Question 1210461
The goal is to find an additional condition that proves $\triangle ABD \cong \triangle ACD$.

## 📐 Congruence Criteria

We are given two facts about $\triangle ABD$ and $\triangle ACD$:

1.  **Shared Side:** $AD$ is common to both triangles, so **$AD = AD$ (Side)**.
2.  **Angle Bisector:** $AD$ bisects $\angle BAC$, which means the angle at the vertex $A$ is divided into two equal angles. Therefore, **$\angle BAD = \angle CAD$ (Angle)**.

So far, we have one **Side** and one **Angle** ($S$ and $A$). To prove congruence, we need one of the following criteria:

* **SAS** (Side-Angle-Side): We need the other side adjacent to the known angle: $AB = AC$.
* **AAS** (Angle-Angle-Side): We need another angle and a non-included side: $\angle ABD = \angle ACD$ (and $AD=AD$) or $\angle ADB = \angle ADC$ (and $AD=AD$).
* **ASA** (Angle-Side-Angle): We need another angle, and the side must be the included side: $\angle ADB = \angle ADC$.

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## 🔎 Evaluating the Options

The options introduce segments and angles from other potential figures ($PQ$, $BPQ$, $PAQ$), which are not necessarily part of $\triangle ABD$ or $\triangle ACD$. We must assume that the non-$A, B, C, D$ elements in the options are meant to relate to the required conditions through substitution or equivalence.

### (a) $BC = AD$

**False.** This compares the side $BC$ of the entire triangle to the segment $AD$. This is not one of the required congruence conditions.

### (b) $AB = PQ$

**True (Potentially).** The required condition for **SAS** congruence is **$AB = AC$**.
If the condition given is **$AB = PQ$**, and we assume the intended required condition **$AB = AC$** can be satisfied by assuming $PQ$ is equal to $AC$ (i.e., $PQ = AC$), then $AB = AC$.
More simply, if we assume the statement means $AB = AC$, then **SAS** is satisfied: $\mathbf{AB = AC}$ (Side), $\mathbf{\angle BAD = \angle CAD}$ (Angle), $\mathbf{AD = AD}$ (Side).

### (c) $\angle ABC = \angle BPQ$

**True (Potentially).** The required condition for **AAS** congruence is $\angle ABD = \angle ACD$, which is the same as **$\angle ABC = \angle ACB$**.
If the condition given is $\mathbf{\angle ABC = \angle ACB}$, then **AAS** is satisfied: $\mathbf{\angle ABD = \angle ACD}$ (Angle), $\mathbf{\angle BAD = \angle CAD}$ (Angle), $\mathbf{AD = AD}$ (Non-included Side).
Assuming the statement intends to mean $\angle ABC = \angle ACB$.

### (d) $\angle BCD + \angle PAQ = 90^\circ$

**False.** This introduces a numerical constraint on two angles and does not provide an equality of sides or angles between $\triangle ABD$ and $\triangle ACD$.

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## ⭐️ Conclusion (Standard Geometry Test)

In a typical geometry problem of this nature, the options are testing whether the student knows the necessary congruence criteria, even if the distracting labels ($PQ$, $BPQ$) are used.

The two easiest ways to prove congruence, given the existing $\mathbf{S}$ and $\mathbf{A}$, are by establishing the **SAS** or **AAS** criteria.

1.  **SAS requires $AB = AC$**
2.  **AAS requires $\angle ABC = \angle ACB$**

Since $\triangle ABD$ and $\triangle ACD$ are the focus, we must select the options that are equivalent to $AB=AC$ or $\angle ABC = \angle ACB$.

* (b) is equivalent to $AB = AC$.
* (c) is equivalent to $\angle ABC = \angle ACB$.

Therefore, both (b) and (c) could be the necessary additional condition, assuming the external angles/segments ($PQ, BPQ$) are equivalent to the required internal ones ($AC, ACB$).

The statements that could be that condition are **(b)** and **(c)**.