Question
Prove that altitudes of a triangle are concurrent
✓
Answer
Consider $\triangle ABC$.
Let $AP \perp BC$ and $BQ \perp AC$.
Let $AP$ and $BQ$ intersect at $O$.
Join $O C$ and extend $O C$ to meet $A B$ at $R$.
To prove that $C R$ is also the altitude of $\triangle A B C$.
i.e., to prove that $C R \perp A B$
Let $\overline{ a }, \overline{ b }, \overline{ c }$ be the position vectors of the points $A , B , C$ respectively.
Consider $\overline{ AP } \perp \overline{ BC }$
$\therefore \overline{ AO } \perp \overline{ BC }$
$\therefore \overline{ AO } \cdot \overline{ BC }=0$
$\therefore-\overline{ a } \cdot(\overline{ c }-\overline{ b })=0 \quad \ldots \ldots .[\because \overline{ AO }=-\overline{ OA }]$
$\therefore \overline{ a } \cdot \overline{ c }-\overline{ a } \cdot \overline{ b }=0 \quad \ldots \ldots . .( i )$
Now, $\overline{ BQ } \perp \overline{ AC }$
$\therefore \overline{ BO } \perp \overline{ AC }$
$\therefore \overline{ BO } \cdot \overline{ AC }=0$
$\therefore-\overline{ b } \cdot(\overline{ c }-\overline{ a })=0 \ldots \ldots[\because \overline{ BO }=-\overline{ OB }]$
$\therefore \overline{ b } \cdot \overline{ c }-\overline{ b } \cdot \overline{ a }=0 \ldots \ldots . .( ii )$
Comparing equations (i) and (ii), we get
$\therefore \overline{ a } \cdot \overline{ c }-\overline{ a } \cdot \overline{ b }=\overline{ b } \cdot c -\overline{ b } \cdot \overline{ a }$
$\therefore \overline{ a } \cdot \overline{ c }=\overline{ b } \cdot \overline{ c }$
$\therefore \overline{ a } \cdot \overline{ c }-\overline{ b } \cdot \overline{ c }=0$
$\therefore \overline{ c } \cdot(\overline{ a }-\overline{ b })=0$
$\therefore-\overline{ c } \cdot(\overline{ a }-\overline{ b })=0$
$\therefore \overline{ CO } \perp \overline{ BA }$
$\therefore \overline{ CR } \perp \overline{ BA }$
$\therefore CR \perp BA$
$\therefore C R$ is also the altitude of $\triangle A B C$.
$\therefore AP , BQ , CR$ intersect at $O$.
$\therefore$ All three altitudes of $\triangle ABC$ intersect at a common point.
Thus, the altitudes of a triangle are concurrent.
Let $AP \perp BC$ and $BQ \perp AC$.
Let $AP$ and $BQ$ intersect at $O$.
Join $O C$ and extend $O C$ to meet $A B$ at $R$.
To prove that $C R$ is also the altitude of $\triangle A B C$.
i.e., to prove that $C R \perp A B$
Let $\overline{ a }, \overline{ b }, \overline{ c }$ be the position vectors of the points $A , B , C$ respectively.
Consider $\overline{ AP } \perp \overline{ BC }$
$\therefore \overline{ AO } \perp \overline{ BC }$
$\therefore \overline{ AO } \cdot \overline{ BC }=0$
$\therefore-\overline{ a } \cdot(\overline{ c }-\overline{ b })=0 \quad \ldots \ldots .[\because \overline{ AO }=-\overline{ OA }]$
$\therefore \overline{ a } \cdot \overline{ c }-\overline{ a } \cdot \overline{ b }=0 \quad \ldots \ldots . .( i )$
Now, $\overline{ BQ } \perp \overline{ AC }$
$\therefore \overline{ BO } \perp \overline{ AC }$
$\therefore \overline{ BO } \cdot \overline{ AC }=0$
$\therefore-\overline{ b } \cdot(\overline{ c }-\overline{ a })=0 \ldots \ldots[\because \overline{ BO }=-\overline{ OB }]$
$\therefore \overline{ b } \cdot \overline{ c }-\overline{ b } \cdot \overline{ a }=0 \ldots \ldots . .( ii )$
Comparing equations (i) and (ii), we get
$\therefore \overline{ a } \cdot \overline{ c }-\overline{ a } \cdot \overline{ b }=\overline{ b } \cdot c -\overline{ b } \cdot \overline{ a }$
$\therefore \overline{ a } \cdot \overline{ c }=\overline{ b } \cdot \overline{ c }$
$\therefore \overline{ a } \cdot \overline{ c }-\overline{ b } \cdot \overline{ c }=0$
$\therefore \overline{ c } \cdot(\overline{ a }-\overline{ b })=0$
$\therefore-\overline{ c } \cdot(\overline{ a }-\overline{ b })=0$
$\therefore \overline{ CO } \perp \overline{ BA }$
$\therefore \overline{ CR } \perp \overline{ BA }$
$\therefore CR \perp BA$
$\therefore C R$ is also the altitude of $\triangle A B C$.
$\therefore AP , BQ , CR$ intersect at $O$.
$\therefore$ All three altitudes of $\triangle ABC$ intersect at a common point.
Thus, the altitudes of a triangle are concurrent.
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