The area of an isosceles triangle $\triangle ABC$ with base $AB=2c$ is $S.$ Find the leg of the triangle.
Let $AC=BC$ and $CH$ and $BP$ be the altitudes through $C$ and $B$, respectively.
I am not sure how to approach the problem. The area of the triangle is $S_{\triangle ABC}=\dfrac{AB.CH}{2}=\dfrac{AC.BP}{2}.$ What next?
Best Answer
Remember the definition of a right angle and the perpendicular bisector is that the perpendicular bisector of $AB$ is all the points $x$ so that $Ax = Bx$ and a right angle with vertex $M$ is one so that if $A,M,B$ are colinear with $AM = BM$ and $M$ between $A$ and $B$ then $\angle AMC$ is a right angle if and only if $AC = BC$.
So as the legs, $AC$ and $BC$ of an isosceles triangle are equal, then $C$ is on the perpendicular bisector of $AB$ and if $M$ is the midpoint of $AB$ then $\triangle AMC$ and $\triangle BMC$ are two congruent right triangles.
Use the Pythogorean Theorem to find $AC = BC$ via
$AC^2 = AM^2 + MC^2$
What is $AM$?
What is $MC$?
So what is $AC$?