Geometry prove isosceles triangle inside a square

Question

I need to prove that there’s an isosceles triangle in a rectangle, dońt know how to do this.

Answer 1

a square by definition has 4 equal sides and 4 equal angles, each = 90 degrees

a triangle with base = one side of the square, call it "b" with length = b

b has end points which are two vertices of the square

construct a line from each of the two vertices to the midpoint of the opposite side

those two lines, plus the base form a triangle with 2 equal sides, making it an isoscelese triangle.

or if you have a rectangle that's not a square, then still use one side of the rectangle as the base of the triangle. construct 2 lines from the vertices of the base to the midpoint of the opposite side. Those 2 lines plus the base form an isoscelese triangle.

Maybe plot them on a graph, with one vertex of the square as the origin, the point (0,0).

Let the base = the length from (0,0) to (b,0), two lower vertices of the square

draw lines from (0,0) to (b/2, b) and from (b,0) to (b/2, b)

(b/2,b) is the midpoint of the side opposite and parallel to the base

those two lines are equal, and when combined with the base, they form an isosceles triangle.

use the pythagorean theorem

from (0,0) to (b/2,b) is a distance = square root of (b^2 + b^2/4)

from (b,0) to (b/2,b) is a distance of square root of (b^2 + b^2/4)

the two lines are equal,

contruct a triangle using those two lines with the base

That forms a 3 sided polygon, a triangle with two equal sides making it an isoscelese triangle.