Question
Prove that the altitudes of a triangle are concurrent.
✓
Answer
Let A, B and C be the vertices of a triangle
Let $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$ be the altitudes of the triangle $\mathrm{ABC}$, therefore $\mathrm{AD} \perp \mathrm{BC}, \mathrm{BE} \perp \mathrm{AC}, \mathrm{CF} \perp \mathrm{AB}$.
Let $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}, \bar{f}$ be the position vectors of $A, B, C, D, E, F$ respectively. Let $\mathrm{P}$ be the point of intersection of the altitudes $A D$ and $B E$ with $\bar{p}$ as the position vector.
Therefore, $\overline{A P} \perp \overline{B C}, \overline{B P} \perp \overline{A C}$
To show that the altitudes $A D, B E$ and $C F$ are concurrent, it is sufficient to show that the altitude $C F$ passes through the point $P$. We will have to prove that $\overline{C F}$ and $\overline{C P}$ are collinear vectors. This can be achieved by showing $\overline{C P} \perp \overline{A B}$
Let $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$ be the altitudes of the triangle $\mathrm{ABC}$, therefore $\mathrm{AD} \perp \mathrm{BC}, \mathrm{BE} \perp \mathrm{AC}, \mathrm{CF} \perp \mathrm{AB}$.
Let $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}, \bar{f}$ be the position vectors of $A, B, C, D, E, F$ respectively. Let $\mathrm{P}$ be the point of intersection of the altitudes $A D$ and $B E$ with $\bar{p}$ as the position vector.
Therefore, $\overline{A P} \perp \overline{B C}, \overline{B P} \perp \overline{A C}$
To show that the altitudes $A D, B E$ and $C F$ are concurrent, it is sufficient to show that the altitude $C F$ passes through the point $P$. We will have to prove that $\overline{C F}$ and $\overline{C P}$ are collinear vectors. This can be achieved by showing $\overline{C P} \perp \overline{A B}$
Now from (1) we have
$
\begin{aligned}
& \overline{A P} \perp \overline{B C} \quad \text { and } \overline{B P} \perp \overline{A C} \\
& \overline{A P} \perp \overline{B C}=0 \quad \text { and } \overline{B P} \perp \overline{A C}=0 \\
\therefore & (\bar{p}-\bar{a}) \cdot(\bar{c}-\bar{b})=0 \text { and }(\bar{p}-\bar{b}) \cdot(\bar{c}-\bar{a})=0 \\
\therefore & \bar{p} \cdot \bar{c}-\bar{p} \cdot \bar{b}-\bar{a} \cdot \bar{c}+\bar{a} \cdot \bar{b}=0 \\
& \bar{p} \cdot \bar{c}-\bar{p} \cdot \bar{a}-\bar{b} \cdot \bar{c}+\bar{b} \cdot \bar{a}=0
\end{aligned}
$
Therefore, subtracting equation (2) from equation (3), we get
$
\begin{array}{ll}
& -\bar{p} \cdot \bar{a}+\bar{p} \cdot \bar{b}-\bar{b} \cdot \bar{c}+\bar{a} \cdot \bar{c}=0 \quad(\text { Since } \bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{a}) \\
\therefore \quad & \bar{p}(\bar{b}-\bar{a})-\bar{c}(\bar{b}-\bar{a})=0 \\
\therefore & (\bar{p}-\bar{c}) \cdot(\bar{b}-\bar{a})=0 \\
\therefore & \overline{C P} \cdot \overline{A B}=0 \\
\therefore & \overline{C P} \perp \overline{A B}
\end{array}
$
Hence the proof.
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