Let $ H $ be the intersection point of the heights of the acute-angled triangle $ ABC $. From points $ A $ and $ C $, tangents $ AK $ and $ CT $ are drawn to the circle drawn on the segment $ BH $ as on the diameter. Let $ 15 $ and $ 19 $ be the lengths of these tangents. What is the square of smallest possible $ AC $ side length?
It can be easy for me if it is with value. As It is an acute triangle, can't find the way. Please help me with a solution to go with this type in the future.
My Simple Answer:
$AK^2 + AC^2 = CT^2$
$15^2 + AC^2 = 19^2$
$AC^2 = 361 - 225$
$AC^2 = 136$
Let the feet of perpendiculars from point $A$, $B$ and $C$ on the respective opposite sides be $D$, $E$ and $F$ respectively. The circle with diameter $BH$ will pass through points $D$ and $F$. Let the tangents from $A$ and $C$ to this circle be $AS$ and $CT$ respectively.
Considering the power of points $A$ and $C$ with respect to this circle gives,
$AS^{2}=AF\cdot AB$
$CT^{2}=CD\cdot BC$
Now, observe that, $CD\cdot BC=CE\cdot AC$ because $\triangle ACD\sim \triangle BCE$. Similarly, $AF\cdot AB=AE\cdot AC$ because $\triangle CAF\sim \triangle BAE$.
$\Rightarrow AS^{2}=AF\cdot AB=AE\cdot AC$ and similarly $CT^{2}=CE\cdot AC$.
Adding these two equations gives, $AC^{2}=\boxed {AS^{2}+CT^{2}}$
Now plugging in the lengths of the tangents will give the value of $AC^{2}$.