Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapeziu

Q: Wait, why do we reject the case where $k=1$?

That is a fantastic observation! The problem explicitly states that the trapezium is not a parallelogram. If $k=1$, then $\vec{DC} = \vec{AB}$, which means $DC$ is equal and parallel to $AB$. By definition, this makes the figure a parallelogram. In a parallelogram, the diagonals bisect each other, and the midpoints of the parallel sides would coincide with the intersection point, making the line segment $M_1M_2$ degenerate or undefined in the context of this specific proof.

Q: Why can't we just use coordinate geometry with arbitrary points?

You absolutely can! However, coordinate geometry often forces you to choose specific coordinates (like $(0,0), (a,0), (b,c), (d,c)$), which can lead to messy algebraic expressions. Vector methods are coordinate-independent and elegant; they allow us to work with the intrinsic geometric properties of the trapezium without getting bogged down in specific coordinate values.

Q: How do I know if I should use the section formula or just linear combinations?

They are actually two sides of the same coin! The section formula $\vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n}$ is just a specific case of a linear combination $\vec{p} = (1-t)\vec{a} + t\vec{b}$ where the sum of the coefficients is 1. Using the linear combination form is often faster because it directly relates to the concept of collinearity.

Question

Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a trapezium, looking at its diagonals. They cross at a point $P$, a silent intersection in the middle of the shape. It feels like a random point, but geometry is rarely random.

There is a hidden order here, a beautiful symmetry that connects the intersection of the diagonals to the midpoints of the parallel sides. Let's uncover this together using the power of vectors.

To begin, we need a language to describe our shape. Let's place vertex $A$ at the origin, $\vec{0}$. We define the vectors $\vec{AB} = \vec{b}$ and $\vec{AD} = \vec{d}$.

Because $AB$ is parallel to $DC$, we know that $\vec{DC}$ must be a scalar multiple of $\vec{b}$. Let's call this scalar $k$, so $\vec{DC} = k\vec{b}$. Since we are told this is not a parallelogram, we know $k 
eq 1$.

### The Diagonals

Now, let's trace the paths of the diagonals. To get from $A$ to $C$, we can travel along $AD$ and then $DC$. Using the triangle law of vector addition, we get:

$$\vec{AC} = \vec{AD} + \vec{DC} = \vec{d} + k\vec{b}$$

For the second diagonal, $BD$, we travel from $B$ to $A$ and then from $A$ to $D$. This gives us:

$$\vec{BD} = \vec{AD} - \vec{AB} = \vec{d} - \vec{b}$$

These two vectors, $\vec{AC}$ and $\vec{BD}$, are the highways upon which our intersection point $P$ resides.

### The Algebraic Dance

Let $P$ be the intersection point. Since $P$ lies on $AC$, its position vector $\vec{p}$ must be a scalar multiple of $\vec{AC}$. We can write $\vec{p} = s(\vec{d} + k\vec{b})$ for some scalar $s$.

But $P$ also lies on $BD$. Using the section formula, we can express $\vec{p}$ as a linear combination of $\vec{b}$ and $\vec{d}$:

$$\vec{p} = (1-t)\vec{b} + t\vec{d}$$

Now, we have two expressions for the same point $\vec{p}$. Because $\vec{b}$ and $\vec{d}$ are non-collinear, they form a basis for our 2D space. This means we can equate the coefficients of $\vec{b}$ and $\vec{d}$ independently.

Equating the coefficients of $\vec{d}$ gives $s = t$. Equating the coefficients of $\vec{b}$ gives $sk = 1 - t$. Substituting $s = t$ into the second equation, we find $tk = 1 - t$, which simplifies to $t(k+1) = 1$, or:

$$t = \frac{1}{k+1}$$

### The Grand Unification

Finally, let's look at the midpoints $M_1$ and $M_2$ of the parallel sides $AB$ and $CD$. Their position vectors are:

$$\vec{m_1} = \frac{1}{2}\vec{b}$$

$$\vec{m_2} = \vec{d} + \frac{k}{2}\vec{b}$$

Any point on the line segment $M_1M_2$ can be written as $\vec{r} = (1-u)\vec{m_1} + u\vec{m_2}$. If we substitute $u = \frac{1}{k+1}$ into this expression, the algebra collapses, and we arrive exactly at our expression for $\vec{p}$.

We have proven it! The intersection point $P$ lies on the line joining the midpoints. It is a beautiful, elegant result that reveals the deep harmony within the geometry of a trapezium.

Animated Solution for Mathematics - Vector Algebra: Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)

Visualized Solution

Set up the Vector Framework

Express Diagonals $A C$ and $B D$

Locate the Intersection Point $P$

Express $P$ along Diagonal $B D$

Equate Coefficients of $b$ and $d$

Solve for the Position Vector $p$

Locate Midpoints $M_{1}$ and $M_{2}$

Verify $P$ lies on Line $M_{1} M_{2}$

Conclusion & Geometric Insight

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Diagonals

The Algebraic Dance

The Grand Unification

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Animated Solution for Mathematics - Vector Algebra: Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)

Visualized Solution

Set up the Vector Framework

Express Diagonals .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BD

Locate the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Express .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P along Diagonal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BD

Equate Coefficients of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Solve for the Position Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p​

Locate Midpoints .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2​

Verify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P lies on Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M1​M2​