analytic geometrycoordinate systemsgeometry
How to show that a line pass through a point?
Two fixed straight line $OX$ and $OY$ are cut by a variable line at the points $A$ and $B$ respectively and $P$ and $Q$ are the feet of the perpendiculars drawn from $A$ and $B$ upon the lines $OBY$ and $OAX$ show that,if $AB$ pass through a fixed point,then $PQ$ will pass through a fixed point.
Well, recall the length of a perpendicular from point $(x_{1},y_{1})$ to the line $ax+by+c=0$ is equal to the distance between them and is thus given by $$\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^2+b^2}}$$
So the length of a perpendicular from point $(2a,2a)$ on the line $mx-y+a=0$ would be given by $$\dfrac{|2ma-a|}{\sqrt{m^2+1}}=a$$ So squaring and dividing by $a^2$ we have $$\dfrac{4m^2-4m+1}{m^2+1}=1 \implies m=0 \text{ or } m=\dfrac{4}{3}$$
Since $OA' \times OA = OP^2$ and $\angle A'OP = \angle POA$, $\triangle OA'P \sim \triangle OPA$. Then,
$\angle OPA' = \angle OAP = 90° - \angle PAY = 90° - \angle PXY$
$= 90° - \angle XPZ - \angle XZP = 90° - \frac{\angle POZ}{2} = \angle OPZ = \angle OZP$
It implies $PA'Z$ is a straight line and $OPAZ$ is a cyclic quadrilateral.
Let $M$ be the other intersection point of $OA$ and the circle $AXPY$. Joint $PM$ and $PY$.
$\angle PMA = \angle PYA = \angle YAM = 90°$
$\Rightarrow AMPY$ is a rectangle
$\Rightarrow PM$ is equal and parallel to $YA$
Let $P'$ be the other intersection point of $PM$ and the circle $PXZ$. Also, delete the line which is parallel to $YA$ and passes through $A'$.
$\triangle P'AM \cong \triangle PAM$
$\Rightarrow \angle P'AM = \angle PAO = \angle PZO = \angle ZPO = \angle ZAO$
$\Rightarrow ZP'A$ is a straight line.
When $MA \lt A'A$ and $OM \lt OP$,
Joint $A'P'$, $A'X$, $P'X$ and $MX$.
$\angle A'P'X = \angle PP'X + \angle PP'A' = \angle PP'X + \angle P'PZ = \angle PP'X + \angle P'XC = \angle P'CZ = \angle AYZ = \angle AMX$
$\Rightarrow A'MXP'$ is a cyclic quadrilateral.
Let $N$ be the other intersection point of $ZY$ and the circle $A'XP'M$. Joint $A'N$.
$\Rightarrow \angle A'NH = \angle XMA = \angle AYH$
$\Rightarrow A'N$ is parallel to $YA$.
$\Rightarrow \angle NHA' = \angle YHA$ and $NA' = P'M = PM = YA$
$\Rightarrow \triangle NA'H \cong \triangle YAH$
$\Rightarrow A'H = AH$
$\Rightarrow$ Since $A'$ and $A$ are fixed points, $H$ is a fixed point.
When $MA \ge A'A$ and $OM \lt OP$,
Let $C$ be the intersection point of $PP'$ and $YX$. Joint $A'P'$, $A'X$, $P'X$ and $MX$.
$\angle A'P'X = \angle PP'X - \angle PP'A' = \angle PP'X - \angle P'PZ = \angle PP'X - \angle P'XC = \angle P'CZ = \angle AYZ = \angle AMX$
$\Rightarrow A'XP'M$ is a cyclic quadrilateral.
Let $N$ be the other intersection point of $ZY$ and the circle $A'XP'M$. Joint $A'N$.
$\Rightarrow \angle A'NH = \angle A'MX = \angle AYH$
$\Rightarrow A'N$ is parallel to $YA$.
$\Rightarrow \angle NHA' = \angle YHA$ and $NA' = P'M = PM = YA$
$\Rightarrow \triangle NA'H \cong \triangle YAH$
$\Rightarrow A'H = AH$
$\Rightarrow$ Since $A'$ and $A$ are fixed points, $H$ is a fixed point.
Best Answer
Say $AB$ always passes through $N$. Let $N$ be the point such that $\angle XON=\angle MOY$ and $\displaystyle\frac{ON}{OM}=\cos\theta$, where $\theta=\angle XOY$; we claim that $N$ always lies on line $PQ$. Indeed, since $OQ=OB\cos \theta$ and $OP=OA\cos \theta$, $\triangle OPQ\sim\triangle OAB$ with scale factor $\cos \theta$. Because the corresponding angles $\angle QON$ and $\angle MOB$ are equal, $M$ corresponds to $N$ in $OAB$'s similar triangle, so $N$ lies on $PQ$ always.