Question: User: 4.11.R Triangle Congruence: CPCTC, Part 2, Homework
Question 1 of 6
Given that $AP \perp AB$, $BQ \perp AB$, and $AP \cong BQ$, prove $O$ is the midpoint of $AB$ and $PQ$.
Graph shape: a horizontal segment $AB$ with point $O$ near the middle, a vertical segment $AP$ rising upward from $A$ on the left, a vertical segment $BQ$ dropping downward from $B$ on the right, and a diagonal segment $PQ$ crossing the horizontal segment at $O$ from $P$ to $Q$; right-angle markers at $A$ and $B$, and congruence tick marks on $AP$ and $BQ$, position_hint \in {center}
1. **State the problem:**
Given $AP \perp AB$, $BQ \perp AB$, and $AP \cong BQ$, prove that $O$ is the midpoint of both $AB$ and $PQ$.
2. **Understand the setup:**
- $AB$ is a horizontal segment.
- $AP$ is perpendicular to $AB$ at $A$ and extends upward.
- $BQ$ is perpendicular to $AB$ at $B$ and extends downward.
- $AP \cong BQ$ means the lengths of $AP$ and $BQ$ are equal.
- $PQ$ is a segment connecting points $P$ and $Q$.
- $O$ lies on $AB$ where $PQ$ crosses $AB$.
3. **Goal:**
Show $O$ is the midpoint of $AB$ and $PQ$, meaning:
$$ AO = OB \quad \text{and} \quad PO = OQ $$
4. **Use right triangles:**
Triangles $APO$ and $BQO$ are right triangles because $AP$ and $BQ$ are perpendicular to $AB$.
5. **Identify congruent parts:**
- $AP \cong BQ$ (given)
- $AO \cong OB$ (to be proven)
- $\angle AP O$ and $\angle BQ O$ are right angles (since $AP \perp AB$ and $BQ \perp AB$)
6. **Prove triangles $APO$ and $BQO$ are congruent:**
- $AP = BQ$ (given)
- $\angle APO = \angle BQO = 90^\circ$
- $OP$ is common to both triangles
By the Hypotenuse-Leg (HL) theorem for right triangles, triangles $APO$ and $BQO$ are congruent:
$$ \triangle APO \cong \triangle BQO $$
7. **Use CPCTC (Corresponding Parts of Congruent Triangles are Congruent):**
From congruence, corresponding segments are equal:
$$ AO = OB \quad \text{and} \quad PO = OQ $$
8. **Conclusion:**
Since $AO = OB$, $O$ is the midpoint of $AB$.
Since $PO = OQ$, $O$ is the midpoint of $PQ$.
Therefore, $O$ is the midpoint of both $AB$ and $PQ$ as required.