
5. Suppose $P$ is an interior point of a triangle $A B C$ such that the ratios

$$
\frac{d(A, B C)}{d(P, B C)}, \quad \frac{d(B, C A)}{d(P, C A)}, \quad \frac{d(C, A B)}{d(P, A B)}
$$

are all equal. Find the common value of these ratios. [Here $d(X, Y Z)$ denotes the perpendicular distance from a point $X$ to the line $Y Z$.]


Solution: Let $A P, B P, C P$ when extended, meet the sides $B C, C A, A B$ in $D, E, F$ respectively. Draw $A K, P L$ perpendicular to $B C$ with $K, L$ on $B C$.(See Fig. 3.)

![](https://cdn.mathpix.com/cropped/2024_06_05_383b92723c75cbf4d4c9g-7.jpg?height=549&width=808&top_left_y=248&top_left_x=681)

Fig. 3.
Fig. 4.

Now

$$
\frac{d(A, B C)}{d(P, B C)}=\frac{A K}{P L}=\frac{A D}{P D}
$$

Similarly,

$$
\frac{d(B, C A)}{d(P, C A)}=\frac{B E}{P E} \quad \text { and } \frac{d(C, A B)}{d(P, A B)}=\frac{C F}{P F}
$$

So, we obtain

$$
\frac{A D}{P D}=\frac{B E}{P E}=\frac{C F}{P F}, \quad \text { and hence } \quad \frac{A P}{P D}=\frac{B P}{P E}=\frac{C P}{P F}
$$

From $\frac{A P}{P D}=\frac{B P}{P E}$ and $\angle A P B=\angle D P E$, it follows that triangles $A P B$ and $D P E$ are similar. So $\angle A B P=\angle D E P$ and hence $A B$ is parallel to $D E$.

Similarly, $B C$ is parallel to $E F$ and $C A$ is parallel to $D F$. Using these we obtain

$$
\frac{B D}{D C}=\frac{A E}{E C}=\frac{A F}{F B}=\frac{D C}{B D}
$$

whence $B D^{2}=C D^{2}$ or which is same as $B D=C D$. Thus $D$ is the midpoint of $B C$. Similarly $E, F$ are the midpoints of $C A$ and $A B$ respectively.

We infer that $A D, B E, C F$ are indeed the medians of the triangle $A B C$ and hence $P$ is the centroid of the triangle. So

$$
\frac{A D}{P D}=\frac{B E}{P E}=\frac{C F}{P F}=3
$$

and consequently each of the given ratios is also equal to 3 .

## $\underline{\text { Aliter }}$

Let $A B C$, the given triangle be placed in the $x y$-plane so that $B=(0,0), C=(a, 0)$ (on the $x$-axis). (See Fig. 4.)

Let $A=(h, k)$ and $P=(u, v)$. Clearly $d(A, B C)=k$ and $d(P, B C)=v$, so that

$$
\frac{d(A, B C)}{d(P, B C)}=\frac{k}{v}
$$

The equation to $C A$ is $k x-(h-a) y-k a=0$. So

$$
\begin{aligned}
\frac{d(B, C A)}{d(P, C A)} & =\frac{-k a}{\sqrt{k^{2}+(h-a)^{2}}} / \frac{(k u-(h-a) v-k a)}{\sqrt{k^{2}+(h-a)^{2}}} \\
& =\frac{-k a}{k u-(h-a) v-k a}
\end{aligned}
$$

Again the equation to $A B$ is $k x-h y=0$. Therefore

$$
\begin{aligned}
\frac{d(C, A B)}{d(P, A B)} & =\frac{k a}{\sqrt{h^{2}+k^{2}}} / \frac{(k u-h v)}{\sqrt{h^{2}+k^{2}}} \\
& =\frac{k a}{k u-h v}
\end{aligned}
$$

From the equality of these ratios, we get

$$
\frac{k}{v}=\frac{-k a}{k u-(h-a) v-k a}=\frac{k a}{k u-h v}
$$

The equality of the first and third ratios gives $k u-(h+a) v=0$. Similarly the equality of second and third ratios gives $2 k u-(2 h-a) v=k a$. Solving for $u$ and $v$, we get

$$
u=\frac{h+a}{3}, \quad v=\frac{k}{3}
$$

Thus $P$ is the centroid of the triangle and each of the ratios is equal to $\frac{k}{v}=3$.
