Animated Solution for Mathematics - Vector Algebra: From a point $O$ inside a triangle $ABC$, perpendiculars $OD,OE,OF$ are drawn to the sides $BC,CA,AB$ respectively. Prove that the perpendiculars from $A,B,C$ to the sides $EF,FD,DE$ are concurrent.

A Vector Odyssey

Welcome, fellow traveler of the JEE Advanced path. Today, we are not just solving a geometry problem; we are witnessing the elegance of vector algebra.

Often, when we see a problem involving triangles and perpendiculars, our instinct is to reach for Euclidean geometry—chasing angles, looking for cyclic quadrilaterals, or hunting for similar triangles. But today, we are going to transcend that. We are going to use the power of vectors to dismantle this problem from the inside out.

The Vector Setup

Imagine you are standing at point $O$ inside triangle $ABC$. This point is our command center. By setting $O$ as our origin, we simplify the entire universe of this problem.

The vertices $A,B,$ and $C$ are now defined by their position vectors $\vec{a},\vec{b},$ and $\vec{c}$. From our command center, we drop perpendiculars to the sides. Let the feet of these perpendiculars be $D,E,$ and $F$, with position vectors $\vec{d},\vec{e},$ and $\vec{f}$.

Now, here is the secret: what does a perpendicular mean in the language of vectors? It means the dot product is zero. Since $OD\perp BC$, we know that:

We can repeat this for the other sides, giving us three foundational equations that will serve as our toolkit throughout this journey:

The Pedal Triangle and the Intersection $H$

Now, connect $D,E,$ and $F$ to form the pedal triangle. The problem asks us to prove that the perpendiculars from $A,B,$ and $C$ to the sides of this new triangle are concurrent.

Let's be strategic. We don't know where all three meet, so let's define point $H$ as the intersection of the first two perpendiculars: the one from $A$ to $EF$ and the one from $B$ to $FD$. Let the position vector of $H$ be $\vec{r}$.

Our mission is now clear: if we can prove that the line $CH$ is perpendicular to $DE$, then $H$ must be the point where all three lines meet. We are essentially forcing the third line to join the party.

The Algebraic Dance

This is where the magic happens. We write the condition for $AH\perp EF$ as:

Similarly, for $BH\perp FD$, we have:

Let's expand these. The first becomes:

The second becomes:

Now, watch the beauty of cancellation. If we add these two equations, the term $-\vec{r}\cdot \vec{f}$ and $+\vec{r}\cdot \vec{f}$ vanish into thin air! We are left with:

The Grand Finale

Remember our initial conditions? We know that $\vec{a}\cdot \vec{f}=\vec{b}\cdot \vec{f}$. Look at our combined equation—we have a $+\vec{a}\cdot \vec{f}$ and a $-\vec{b}\cdot \vec{f}$. They cancel out perfectly!

We are left with:

We need to introduce $\vec{c}$ to prove the final perpendicularity. Using our initial relations, we know $\vec{a}\cdot \vec{e}=\vec{c}\cdot \vec{e}$ and $\vec{b}\cdot \vec{d}=\vec{c}\cdot \vec{d}$. Substituting these in, we get:

Finally, we factorize. Grouping the terms gives us:

This is the moment of truth. $(\vec{r}-\vec{c})$ is the vector $\overrightarrow{CH}$, and $(\vec{e}-\vec{d})$ is the vector $\overrightarrow{DE}$. Their dot product is zero, which means $CH\perp DE$.

The third perpendicular passes through $H$. The proof is complete. You have just conquered a complex geometric problem using nothing but the elegant logic of vectors.