Subfield: Geometry


Question: In the diagram, quadrilateral $A B C D$ has points $M$ and $N$ on $A B$ and $D C$, respectively, with $\frac{A M}{A B}=\frac{N C}{D C}$. Line segments $A N$ and $D M$ intersect at $P$, while $B N$ and $C M$ intersect at $Q$. Prove that the area of quadrilateral $P M Q N$ equals the sum of the areas of $\triangle A P D$ and $\triangle B Q C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4e0d9d101cf875a61e98g-1.jpg?height=358&width=629&top_left_y=1035&top_left_x=1168)

Solution:
We use the notation $|P M Q N|$ to represent the area of quadrilateral $|P M Q N|,|\triangle A P D|$ to represent the area of $\triangle A P D$, and so on.

We want to show that $|P M Q N|=|\triangle A P D|+|\triangle B Q C|$.

This is equivalent to showing

$$
|P M Q N|+|\triangle D P N|+|\triangle C Q N|=|\triangle A P D|+|\triangle D P N|+|\triangle B Q C|+|\triangle C Q N|
$$

which is equivalent to showing

$$
|\triangle D M C|=|\triangle D A N|+|\triangle C B N|
$$

since combining quadrilateral $P M Q N$ with $\triangle D P N$ and $\triangle C Q N$ gives $\triangle D M C$, combining $\triangle A P D$ with $\triangle D P N$ gives $\triangle D A N$, and combining $\triangle B Q C$ with $\triangle C Q N$ gives $\triangle C B N$. Suppose that $D C$ has length $x$ and $D N$ has length $t x$ for some $t$ with $0<t<1$.

Then $N C=D C-D N=x-t x=(1-t) x$.

Suppose also that the height of $A$ above $D C$ is $a$, the height of $B$ above $D C$ is $b$ and the height of $M$ above $D C$ is $m$.

![](https://cdn.mathpix.com/cropped/2023_12_21_36090a78f0d340a667c7g-1.jpg?height=420&width=639&top_left_y=1869&top_left_x=843)

Figure 1

Then $|\triangle D A N|=\frac{1}{2}(t x)(a)$ and $|\triangle C B N|=\frac{1}{2}((1-t) x) b$ so

$$
|\triangle D A N|+|\triangle C B N|=\frac{1}{2}(t x a+(1-t) x b)=\frac{1}{2} x(t a+(1-t) b)
$$

Also, $|\triangle D M C|=\frac{1}{2} x m$.

In order to prove that $|\triangle D M C|=|\triangle D A N|+|\triangle C B N|$, we need to show that $\frac{1}{2} x m$ equals



$\frac{1}{2} x(t a+(1-t) b)$ which is equivalent to showing that $m$ is equal to $t a+(1-t) b$.

In Figure 2, we draw a horizontal line from $A$ to $B G$, meeting $M F$ at $R$ and $B G$ at $S$.

Since $M F$ and $B G$ are vertical and $A R S$ is horizontal, then these line segments are perpendicular.

Since $A E=a, M F=m$ and $B G=b$, then $M R=m-a$ and $B S=b-a$.

![](https://cdn.mathpix.com/cropped/2023_12_21_8e2034ac0213e540ad30g-1.jpg?height=350&width=398&top_left_y=451&top_left_x=969)

Figure 2

Now $\triangle A R M$ is similar to $\triangle A S B$, since each is right-angled and they share a common angle at $A$.

Therefore, $\frac{M R}{B S}=\frac{A M}{A B}=\frac{N C}{D C}$.

Since $M R=m-a$ and $B S=b-a$, then $\frac{M R}{B S}=\frac{m-a}{b-a}$.

Since $\frac{A M}{A B}=\frac{N C}{D C}$, then $\frac{M R}{B S}=\frac{(1-t) x}{x}=1-t$.

Comparing these two expressions, we obtain $\frac{m-a}{b-a}=(1-t)$ or $m-a=(b-a)(1-t)$ or $m=a+b(1-t)+(t-1) a=t a+(1-t) b$, as required.

This concludes the proof, and so $|P M Q N|=|\triangle A P D|+|\triangle B Q C|$, as required.
Let $A M=x$ and $M B=y$. Then $A B=x+y$ and so $\frac{A M}{A B}=\frac{x}{x+y}$.

Let $N C=n x$ for some real number $n$.

Since $\frac{N C}{D C}=\frac{A M}{A B}$, then $\frac{n x}{D C}=\frac{x}{x+y}$ and so $D C=n(x+y)$.

This tells us that $D N=D C-N C=n(x+y)-n x=n y$.

Join $M$ to $N$ and label the areas as shown in the diagram:

![](https://cdn.mathpix.com/cropped/2023_12_21_8e2034ac0213e540ad30g-1.jpg?height=371&width=637&top_left_y=1926&top_left_x=842)

We repeatedly use the fact that triangles with a common height have areas in proportion to the lengths of their bases.

For example, $\triangle M D N$ and $\triangle M N C$ have a common height from line segment to $D C$ to $M$ and so the ratio of their areas equals the ratio of the lengths of their bases.

In other words, $\frac{w+r}{u+v}=\frac{n x}{n y}=\frac{x}{y}$. Thus, $w+r=\frac{x}{y}(u+v)$.



Also, the ratio of the area of $\triangle N A M$ to the area of $\triangle N M B$ equals the ratio of $A M$ to $M B$.

This gives $\frac{k+v}{s+w}=\frac{x}{y}$ or $k+v=\frac{x}{y}(s+w)$.

Next, we join $A$ to $C$ and relabel the areas divided by this new line segment as shown:

![](https://cdn.mathpix.com/cropped/2023_12_21_ba5be69ca99855827d89g-1.jpg?height=366&width=634&top_left_y=430&top_left_x=846)

(The unlabelled triangle adjacent to the one labelled $k_{1}$ has area $k_{2}$ and the unlabelled triangle adjacent to the one labelled $r_{2}$ has area $r_{1}$.)

Consider $\triangle A N C$ and $\triangle A D N$.

As above, the ratio of their areas equals the ratio of their bases.

Thus, $\frac{k_{2}+v_{2}+w_{2}+r_{2}}{z+u}=\frac{n x}{n y}=\frac{x}{y}$, and so $k_{2}+v_{2}+w_{2}+r_{2}=\frac{x}{y}(z+u)$.

Consider $\triangle C A M$ and $\triangle C M B$.

As above, the ratio of their areas equals the ratio of their bases.

Thus, $\frac{k_{1}+v_{1}+w_{1}+r_{1}}{s+t}=\frac{x}{y}$, and so $k_{1}+v_{1}+w_{1}+r_{1}=\frac{x}{y}(s+t)$.

Adding $k_{2}+v_{2}+w_{2}+r_{2}=\frac{x}{y}(z+u)$ and $k_{1}+v_{1}+w_{1}+r_{1}=\frac{x}{y}(s+t)$ gives

$$
\left(k_{1}+k_{2}\right)+\left(v_{1}+v_{2}\right)+\left(w_{1}+w_{2}\right)+\left(r_{1}+r_{2}\right)=\frac{x}{y}(s+t+z+u)
$$

or

$$
k+v+w+r=\frac{x}{y}(s+t+z+u)
$$

Since $k+v=\frac{x}{y}(s+w)$ and $w+r=\frac{x}{y}(u+v)$, then

$$
\frac{x}{y}(s+w)+\frac{x}{y}(u+v)=\frac{x}{y}(s+t+z+u)
$$

which gives

$$
s+w+u+v=s+t+z+u
$$

or

$$
w+v=t+z
$$

But $w+v$ is the area of quadrilateral $P M Q N, z$ is the area of $\triangle A P D$ and $t$ is the area of $\triangle B Q C$. In other words, the area of quadrilateral $P M Q N$ equals the sum of the areas of $\triangle A P D$ and $\triangle P Q C$, as required.

Final answer: 证明题，略