Given Triangle, Any Side, Any Point!

Imgur

In quadrilateral, APDQ is cyclic since both $\angle P = \angle Q = 90°$ and a circle with center O circumscribes it.

The cevian AD must be a diameter = 2r

$\angle QOP$ at the center must be twice the angle $\alpha = \angle BAC$ subtended by PQ on the circumference.

So the length of the desired chord $PQ = L = 2r \sin \alpha$

Since angle $\alpha$ = constant for the given triangle, the length L will be minimum when r is minimized.

This will happen when cevian AD = altitude $h_A$

Similarly, altitudes $h_B$ and $h_C$ from B and C will also give candidates for the desired minima,

But $BC \times h_A = AC \times h_B = AB \times h_C$ = constant = 2 Area

and $\frac{BC}{\sin \alpha} = \frac{AC}{\sin \beta} = \frac{AB}{\sin \gamma}$

Hence $L = h_A \sin \alpha = h_C \sin \beta = h_B \sin \gamma$ and the three candidates for minimum L are all equal!

From Heron's formula area A = $\frac{15\sqrt{7}}{4}$ and $h_A = \frac{5\sqrt{7}}{4}$

$\cos \alpha = \frac{1}{8}$ ; $\sin \alpha = \frac{3\sqrt{7}}{8}$

$L = h_A \sin \alpha = \frac{15 \times 7}{32} = \frac{105}{32}$

Such a beautiful result! Thanks for sharing it @Satyajit Mohanty !

Basically,the question is very simple.However,I liked the final result,so thought of sharing the procedure in which I approached the question.Firstly,DP and DQ are altitudes of triangles ADC and ADB respectively.We know that the sum of the areas of the two triangles is equal to that of triangle ABC.That gives us one equation relating DP and DQ.Secondly,as quadrilateral AQDP is cyclic we have angle PDQ equal to (180-A) degrees.Hence,applying cosine rule in triangle PDQ we get PQ entirely in terms of DP and DQ.Now,from the first equation we had got,we can substitute either DP or DQ in the second equation (the one which we got by applying cosine rule),hence we get length of PQ as a quadratic equation in either length of DP or that of DQ.As we know that a quadratic equation achieves minimum value at the vertex of the parabola,we can find the minimum value of length of PQ.If we solve for a general triangle ABC,then following the same method as above,we get the minimum value of length of PQ as Area of triangle ABC divided by circumradius of triangle ABC.