cx is an altitude in triangle abc. which statements are true? choose two correct answers. $\triangle bcx \cong \triangle acx$ $\triangle cxa \cong \triangle cba$ $\triangle axc \sim \triangle cxb$ $\triangle abc \cong \triangle bxc$ $\triangle acb \sim \triangle axc$

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# Brief Explanations:
1. For $\triangle ACB \sim \triangle AXC$:
   - In $\triangle ACB$ and $\triangle AXC$, $\angle ACB=\angle AXC = 90^{\circ}$ (since $CX$ is an altitude, $\angle AXC = 90^{\circ}$ and $\angle ACB$ is given as a right angle? Wait, looking at the diagram, $\angle ACB$ is a right angle? Wait, the diagram shows $\angle ACB$ with a right - angle mark and $CX\perp AB$ (since $CX$ is an altitude). Also, $\angle A$ is common to both $\triangle ACB$ and $\triangle AXC$. By the AA (Angle - Angle) similarity criterion, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar. $\angle A=\angle A$ and $\angle ACB=\angle AXC = 90^{\circ}$, so $\triangle ACB\sim\triangle AXC$.
2. For $\triangle AXC\sim\triangle CXB$:
   - $\angle AXC=\angle CXB = 90^{\circ}$ (since $CX$ is an altitude). Also, $\angle ACX+\angle B=\angle ACX+\angle ACX = 90^{\circ}$ (because in right - triangle $ACB$, $\angle A+\angle B = 90^{\circ}$ and in right - triangle $AXC$, $\angle A+\angle ACX=90^{\circ}$, so $\angle ACX=\angle B$). So in $\triangle AXC$ and $\triangle CXB$, $\angle AXC=\angle CXB = 90^{\circ}$ and $\angle ACX=\angle B$. By AA similarity criterion, $\triangle AXC\sim\triangle CXB$.

- Let's check the other options:
     - For $\triangle BCX\cong\triangle ACX$: There is no information to suggest that the corresponding sides are equal. Just because they are right - triangles does not mean they are congruent.
     - For $\triangle CXA\cong\triangle CBA$: The triangles have different side lengths (e.g., $CX$ is an altitude, much shorter than $BC$ or $AC$ in general), so they are not congruent.
     - For $\triangle ABC\cong\triangle BXC$: The triangles have different shapes and side lengths, so they are not congruent.

# Answer:
$\triangle ACB \sim \triangle AXC$, $\triangle AXC\sim\triangle CXB$

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