G'day, everyone. I am trying to answer the question:
Let AB be a diameter of circle K with centre O. CHoose a point P exterior to K on the line through A and B and construct the tangents to K through P, Meeting K at X and Y. Let M be the intersection of AB and XY. Prove that the tangents to K are also tangents to both circles with centre A and radius AM and circle B and radius BM.
The context for this question is that I had been attempting this mathematics enrichment tasks for just under 3 months now and it is finally coming to an end, however, the difficulty of the questions have increased and I am beginning to have more difficulty in answering as the topics discussed in the enrichment are some I have not seen before. What I have learnt in order to answer: I have been reviewing some basic circle geometry laws in order to finish this question, the ones typically learnt at my level of mathematics are: 1. if a tangent meets a radius, the point of contact is perpendicular, 2. two tangents from an external point are equal, 3. If a line is drawn from the centre to the exterior point of 2 tangents, the line bisects the angle, 4. the "alternate segment theorem", 5. if 2 external circles have 1 intersection point, the line at the point is a tangent. I believe the question will be formatted around these rules as these are the ones expected to be learnt.
Any help would be appreciated and please let me know if there is anything within the question I need to change in order for it to be better formatted for future users of this site.
Best Answer
What we aim to show here is that points $B$ and $A$ are the respective incentre and excentre opposite point $P$ in $\triangle PXY$.
$\triangle PXY$ is isosceles and $BM$ is the perpendicular bisector of $XY$. Hence, $PB$ bisects $\angle P$.
Also, $\angle MXB=\angle BAX=\angle BXP$ and thus $B$ is the incentre of $\triangle PXY$.
$\angle AXB=\angle AYB=90^{\circ}$ and hence $A$ is the excentre opposite point $P$ in $\triangle PXY$.