A nice Geometric Optimisation Problem - Part 2

This post picks up where we left earlier.

In addition to the centre of circumcircle of $\triangle KDL$, the midpoint of the segment $KL$ also lies in the circle. In fact, this can be proved more easily.

Let $A$ and $C$ be the centers of the respective circles. It can be easily seen that $2\angle CAD=\angle BAD=2\angle BQD=2\angle BKD$ which means $\angle CAD=\angle BKD$. Similarly, we can show that $\angle ACD=\angle BLD$. Hence, we see that $\triangle ACD \sim \triangle KLD$.

If we let $O$ be the midpoint of $AC$ and $P$ be the midpoint of $KL$, then by the similarity of the triangle we established before, we have $\angle DOC=\angle DPL=\angle DPB$. Because $\angle DOC$ is constant, we see that $P$ subtends a constant angle with segment $BD$ and hence lies on a circle containing $B$ and $D$.

Also, note that $\angle BOD=2\angle DOC=2\angle DPB=2\angle DSB$ where $S$ is as shown below. Because the angle subtended by $O$ with segment $BD$ is twice that of $P$, we know that $O$ is the center of circle containing $P$.

As both $J$ and $O$ lie on the perpendicular bisector of $AC$ at the product-maximising-position, segment $DT$ (where $T$ is shown below) is parallel to $AC$. Also, $\angle BDT$ is right angled which makes $BT$ the diameter of $P$'s locus.

Because $\angle NDT$ is right angle, we see that $NT$, which is the perpendicular bisector of $KL$ at the product-maximising position, is the diameter of $\triangle KLD$'s circumcircle.

We know that $\angle KTN=\angle KDN$ and $\angle LDN=\angle LTN$ (angles subtended by the same chord in a circle), we see that $\angle KDB=\angle LDB$.

Therefore, we finally see that at the product-maximising position, $ND$ (or) $BD$ should be the angle bisector $\angle KDB$ which is the relation we were looking for in the previous post.

Hope you enjoyed this. See ya later.

Until then
Yours Aye
Me