Two circles $k_1$ and $k_2$ intersect at points $A$ and $B$. An arbitrary secant through $A$ intersects the circles $k_1$ and $k_2$ at points $M$ and $N$, respectively. An arbitrary secant through $B$ intersects the circles $k_1$ and $k_2$ at points $P$ and $Q$, respectively. Prove that $MP$ is parallel to $NQ$.

1. Problem Description

Two circles $k_1$ and $k_2$ intersect at points $A$ and $B$. An arbitrary secant through $A$ intersects the circles $k_1$ and $k_2$ at points $M$ and $N$, respectively. An arbitrary secant through $B$ intersects the circles $k_1$ and $k_2$ at points $P$ and $Q$, respectively. Prove that $MP$ is parallel to $NQ$.

2. Solution Steps

Let $\angle MBA = \alpha$ and $\angle NBA = \beta$.

Since $A, M, B, P$ lie on circle $k_1$, quadrilateral $AMBP$ is cyclic. Therefore, $\angle AMP = 180^{\circ} - \angle ABP$.

Since $A, N, B, Q$ lie on circle $k_2$, quadrilateral $ANBQ$ is cyclic. Therefore, $\angle ANQ = 180^{\circ} - \angle ABQ$.

Let $MP$ and $NQ$ intersect line $AB$ at $X$ and $Y$ respectively.

We want to prove that $MP \parallel NQ$, which is equivalent to showing $\angle BMX = \angle BNY$, where $X$ and $Y$ are such that $X$ is on line $MP$ and $Y$ is on line $NQ$.

Since $AMBP$ is cyclic, $\angle MPA = \angle MBA = \alpha$.

Then $\angle BMP = \angle AMP - \angle AMB = (180^\circ - \angle ABP) - \angle AMB$.

Similarly, since $ANBQ$ is cyclic, $\angle NQA = \angle NBA = \beta$.

Then $\angle BNQ = \angle ANQ - \angle ANB = (180^\circ - \angle ABQ) - \angle ANB$.

We have $\angle PBM = \angle PBA = \alpha$ and $\angle QBN = \angle QBA = \beta$.

Now, $\angle AMP = 180^\circ - \angle ABP$. Thus $\angle BMP = \angle AMB = 180^\circ - \angle MPA - \angle BAM$

$\angle ANQ = 180^\circ - \angle ABQ$. Thus $\angle BNQ = \angle ANB$

$\angle PBA = \angle PMA$ and $\angle QBA = \angle QNA$.

Since $MP$ intersects $AB$ and $NQ$ intersects $AB$, we need to show that alternate interior angles are equal.

Let $\angle AMP = \theta$. Then $\angle ABP = 180^\circ - \theta$. Thus $\angle MBP = \theta$.

Let $\angle ANQ = \phi$. Then $\angle ABQ = 180^\circ - \phi$. Thus $\angle NBQ = \phi$.

Let the line through A intersecting the two circles at M and N. Let line through B intersecting the two circles at P and Q.

$\angle MBA = \angle MPA$ and $\angle NBA = \angle NQA$

Also $\angle MAB = \angle MPB$ and $\angle NAB = \angle NQB$.

$\angle MPA = \angle MBA$ and $\angle QNA = \angle NBA$.

If we let $\angle BAM = x$, then $\angle BAP = x$. Let $\angle BAN = y$, $\angle BAQ = y$.

Since $MP$ and $NQ$ are transversal lines.

Since the angles subtended by $MA$ and $NA$ at $B$ are $\angle MBA = \alpha$, and $\angle NBA = \beta$. Then $\angle MPA = \alpha$ and $\angle NQA = \beta$.

$\angle BMP = 180 - \angle AMP = 180 - (180 - \alpha) = \alpha$.

$\angle BNQ = 180 - \angle ANQ = 180 - (180 - \beta) = \beta$.

Then $\angle MBP = 180 - \angle MAP$

Since $\angle MPA = \angle MBA$ and $\angle NQA = \angle NBA$, we have $\angle MBA - \angle NBA = \angle MPA - \angle NQA$.

$\angle MBA = \angle MPA$ (angles subtended by arc AM)

$\angle NBA = \angle NQA$ (angles subtended by arc AN)

Let $MP \cap AB = X$ and $NQ \cap AB = Y$.

We need to show $\angle AXM = \angle AYN$.

$\angle AXM = 180 - \angle MXA$.

3. Final Answer

MP is parallel to NQ.