Question

prove that of all the triangles with same base between the same parallel line the isosceles triangle has the least perimeter

Answer 1

Draw line segment ABAB, representing the base. Since the area is fixed, this means that the height hh is fixed. Draw a line LLparallel to ABAB that is hh units above ABAB. We seek a point CC on LL that minimizes the distance AC+CBAC+CB.

Now imagine travelling from AA to CC, but then instead of turning back around to BB, we take the mirrored path (reflected in the line LL) and arrive at a point B′B′, which is hhunits above the line LL (and thus 2h2h units above BB). Notice that CB=CB′CB=CB′. So it remains to find CC such that AC+CB′AC+CB′is minimized.

But the shortest distance between any two points is a straight line! Notice by construction that LL bisects the straight line from AA to B′B′. We conclude that taking CC to be the midpoint (so that the triangle is isosceles) will minimize the perimeter.

Answer 2

Answer:

Proved.

Step-by-step explanation:

Draw line segment ABAB, representing the base. Since the area is fixed, this means that the height hh is fixed. Draw a line LLparallel to ABAB that is hh units above ABAB. We seek a point CC on LL that minimizes the distance AC+CBAC+CB.

Now imagine travelling from AA to CC, but then instead of turning back around to BB, we take the mirrored path (reflected in the line LL) and arrive at a point B′B′, which is hhunits above the line LL (and thus 2h2h units above BB). Notice that CB=CB′CB=CB′. So it remains to find CC such that AC+CB′AC+CB′is minimized.

But the shortest distance between any two points is a straight line! Notice by construction that LL bisects the straight line from AA to B′B′. We conclude that taking CC to be the midpoint (so that the triangle is isosceles) will minimize the perimeter.