Given the base of a triangle and ratio of other $2$ sides. Find the locus of the third vertex
To prove that a geometric shape is the correct locus for a given set of conditions. We have to divide the proof into two stages:
- Proof that all the points that satisfy the given conditions are on the given shape.
- Proof that all the points on the given shape satisfy the given conditions.
I am able to prove that the locus of a point which satisfy the satisfy the given conditions is a circle.
Let $BC$ be the base. And $A$ be the third vertex. We are given $AB:AC$ ratio. Angle Bisectors of $\angle A$ meet $BC$ at $P$ and $Q$. Since $\angle PAQ$ is a right angle. The locus of $A$ is a circle with $PQ$ as a diameter. This is first proof.
I couldn't obtain the solution for second proof. I want to prove that all the points on a circle with $PQ$ as a diameter is such that the ratio of the other two sides is constant that we initialised earlier. We initialised $AB:AC$ to be constant. Let a new point on the circle be $A'$. I want to prove that $A'B:A'C$ is same as $AB:AC$.