Given the base of a triangle and the ratio of the lengths of the other two unequal sides , prove that the vertex lies on a fixed circle .

Question

Can someone please explain to me how to approach this question ? I don't want the solutio

Answer 1

You could try it analytically. Let the base be the segment from $P = (-1,0)$ to $Q = (1,0)$. Then write the equation of the locus of points $Z$ such that the distance from $Z$ to $P$ is $k$ times the distance from $Z$ to $Q$.

The locus when $k=1$ is an interesting "circle".

Answer 2

Hint: Let the two fixed points be $A$ and $B$, and let $C$ be one of the possible third vertices. Consider the intersection of the interior/exterior angle bisectors of $\angle ACB$ with line $AB$.

Answer 3

Outline: Not quite true, if the ratio is $1$ then the locus traced out by the vertices is a line. Let $k\ne 1$ be a positive constant.

Without loss of generality we may assume that the base of the triangle is the line segment joining $(-a,0)$ to $(a,0)$.

Let $(x,y)$ be an arbitrary point.

(i) Write down the distance from $(x,y)$ to $(-a,0)$.

(ii) Write down the distance from $(x,y)$ to $(a,0)$.

(iii) Write down the equation that says the ratio of these is $k$.

(iv) Manipulate.