Question 1166052: Let ABCD be a parallelogram and P a point in its interior. Define points
Q, R and S such that A is the midpoint of PQ, B is the midpoint of QR
and C is the midpoint of RS. Prove that D is the midpoint of SP.
Answer by CPhill(2141)   (Show Source):
You can put this solution on YOUR website! This is a great problem that can be solved elegantly using **vector algebra**.
## 📐 Proof using Vector Algebra
Let the origin be $O$. We will represent the position of each point using its position vector (e.g., the position vector of point $A$ is $\vec{a}$).
### 1. Parallelogram Property
Since $ABCD$ is a parallelogram, the sum of vectors from opposite vertices is equal.
$$\vec{d} - \vec{a} = \vec{c} - \vec{b}$$
Rearranging this gives the fundamental vector property of a parallelogram:
$$\vec{a} + \vec{c} = \vec{b} + \vec{d} \quad \text{(Equation 1)}$$
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### 2. Midpoint Conditions
The problem defines points $Q$, $R$, and $S$ based on midpoint conditions:
* **A is the midpoint of PQ:**
$$\vec{a} = \frac{\vec{p} + \vec{q}}{2} \implies \vec{q} = 2\vec{a} - \vec{p} \quad \text{(Equation 2)}$$
* **B is the midpoint of QR:**
$$\vec{b} = \frac{\vec{q} + \vec{r}}{2} \implies \vec{r} = 2\vec{b} - \vec{q} \quad \text{(Equation 3)}$$
* **C is the midpoint of RS:**
$$\vec{c} = \frac{\vec{r} + \vec{s}}{2} \implies \vec{s} = 2\vec{c} - \vec{r} \quad \text{(Equation 4)}$$
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### 3. Substitution and Simplification
We will substitute Equation 2 into Equation 3, and then the result into Equation 4 to express $\vec{s}$ in terms of $\vec{a}$, $\vec{b}$, $\vec{c}$, and $\vec{p}$.
**Substitute $\vec{q}$ (Eq. 2) into $\vec{r}$ (Eq. 3):**
$$\vec{r} = 2\vec{b} - (2\vec{a} - \vec{p})$$
$$\vec{r} = 2\vec{b} - 2\vec{a} + \vec{p} \quad \text{(Equation 5)}$$
**Substitute $\vec{r}$ (Eq. 5) into $\vec{s}$ (Eq. 4):**
$$\vec{s} = 2\vec{c} - (2\vec{b} - 2\vec{a} + \vec{p})$$
$$\vec{s} = 2\vec{c} - 2\vec{b} + 2\vec{a} - \vec{p}$$
Group the vectors using the parallelogram property (Equation 1).
$$\vec{s} = 2(\vec{a} + \vec{c}) - 2\vec{b} - \vec{p}$$
**Now, substitute $\vec{a} + \vec{c} = \vec{b} + \vec{d}$ (from Eq. 1):**
$$\vec{s} = 2(\vec{b} + \vec{d}) - 2\vec{b} - \vec{p}$$
$$\vec{s} = 2\vec{b} + 2\vec{d} - 2\vec{b} - \vec{p}$$
The $2\vec{b}$ terms cancel out:
$$\vec{s} = 2\vec{d} - \vec{p}$$
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### 4. Conclusion
Rearrange the final expression to check the required midpoint condition:
$$\vec{s} = 2\vec{d} - \vec{p}$$
$$\vec{s} + \vec{p} = 2\vec{d}$$
$$\vec{d} = \frac{\vec{p} + \vec{s}}{2}$$
This is the definition of the midpoint of $SP$. Therefore, **$D$ is the midpoint of $SP$**.
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