Isosceles triangle and scalene triangle

Question

Question:

Given the base and vertical angle of a triangle show that its area is greatest when the triangle is isosceles.

My attempt:

For isosceles triangle (with base given 2x, and vertical angle let z):

Note: $y$ is got by $(180 - z)/2$

I calculated area to be: $$CD^2 * tan(y)$$ by using trigonometry and the fact that the altitude in isosceles triangle from vertex angle bisects the base.

I don't know how I can prove this area to be greatest.

Please help me. Thank you!

Answer 1

The circle is the locus of all triangles with a given base and a given angle opposite to it. If a line is drawn parallel to the base there are two points of intersection. Altitude of these two scalene triangle drawn between intersection points can be increased by drawing the parallel line progressively farther away until it cuts at two coincident points as a tangent when the triangle vertex is situated at apex as coalesced tangent point making it isosceles by symmetry.

Answer 2

The inscribed angle theorem gives a circle as the locus of all triangles with a given base and a given angle opposite that base. Among all possible corners on that circle, the one which maximizes height is the one leading to an isosceles triangle.