From a point O inside a triangle ABC,...

MathematicsScalar (Dot) ProductJEE Advanced 1978Difficult

From a point $O$ inside a triangle $A B C$ , perpendiculars $O D, O E, O F$ are drawn to the sides $B C, C A, A B$ respectively. Prove that the perpendiculars from $A, B, C$ to the sides $E F, F D, D E$ are concurrent.

Visualized Solution (Hindi)

Visualizing the Setup

Let the position vectors of vertices $A, B, C$ be $a, b, c$ relative to point $O$ .
Let the feet of perpendiculars $D, E, F$ have position vectors $d, e, f$ .
We need to prove that perpendiculars from $A$ to $E F$ , $B$ to $F D$ , and $C$ to $D E$ meet at a point $H$ .

Defining Orthogonality Conditions

Since $O D ⊥ B C \Rightarrow d \cdot (b - c) = 0 \Rightarrow d \cdot b = d \cdot c$ ... (1)
Since $O E ⊥ A C \Rightarrow e \cdot (c - a) = 0 \Rightarrow e \cdot c = e \cdot a$ ... (2)
Since $O F ⊥ A B \Rightarrow f \cdot (a - b) = 0 \Rightarrow f \cdot a = f \cdot b$ ... (3)

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

Let $H$ be the intersection of perpendiculars from $A$ to $E F$ and $B$ to $F D$ .
Let the position vector of $H$ be $r$ .
Goal: Prove $C H ⊥ D E$ , i.e., $(r - c) \cdot (e - d) = 0$ .

Setting Up Perpendicularity Equations

$A H ⊥ E F \Rightarrow (r - a) \cdot (e - f) = 0$
$\Rightarrow r \cdot e - r \cdot f - a \cdot e + a \cdot f = 0$ ... (4)
$B H ⊥ F D \Rightarrow (r - b) \cdot (f - d) = 0$
$\Rightarrow r \cdot f - r \cdot d - b \cdot f + b \cdot d = 0$ ... (5)

Combining the Equations

Adding (4) and (5):
$(r \cdot e - r \cdot f - a \cdot e + a \cdot f) + (r \cdot f - r \cdot d - b \cdot f + b \cdot d) = 0$
$r \cdot e - a \cdot e + a \cdot f - r \cdot d - b \cdot f + b \cdot d = 0$

Applying Initial Orthogonality

Using $a \cdot f = b \cdot f$ from (3), these terms cancel.
Substitute $a \cdot e = c \cdot e$ from (2) and $b \cdot d = c \cdot d$ from (1):
$r \cdot e - c \cdot e - r \cdot d + c \cdot d = 0$
$\Rightarrow r \cdot (e - d) - c \cdot (e - d) = 0$

Final Proof of Concurrency

Factorizing the expression:
$(r - c) \cdot (e - d) = 0$
This implies $C H \cdot D E = 0$ , so $C H ⊥ D E$ .
Therefore, the perpendicular from $C$ to $D E$ passes through $H$ .
Hence, the three perpendiculars are concurrent.

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