Properties, Characteristics and proofs: Properties of triangles within a square

The diagonals of the square intersect each other, so the four triangles are isosceles. Moreover, since the diagonals are perpendicular to each other, the diagonals form four right-angled triangles. Therefore, the correct answers are A+C

To answer the question, we first need to recall the properties of a square.

In a square, all sides are equal. Therefore, the lengths of all sides, $AB$, $BD$, $DC$ and $AC$ are equal.
Since we denoted $AC=X$, we can state that:

$AB=BD=DC=AC=X$

Now let's remember another property of a square, which is that in a square all angles are equal to $90$ degrees.
This means that triangle $ACD$ is a right triangle, because angle C (which is part of the square) equals $90$ degrees.

In a right triangle, we can use another tool we have - the Pythagorean theorem.
The Pythagorean theorem allows us, in a right triangle, to find the length of the third side using the other two sides.

It goes like this:

$A²+B²=C²$

where C is the hypotenuse.

Therefore we can substitute:

$AC²+CD²=AD²$

Since we know that $AC=CD=X$, the equation will be:

$X²+X²=AD²$

$2X²=AD²$

$\sqrt{2X²}=AD$

$\sqrt2 \cdot X=AD$

And thus we found the third side of the triangle.

But we're not done yet!

Remember, we were asked to find the perimeter of triangle ACD,
Remember, the perimeter of a triangle is the sum of its sides.

$AC+CD+AD= perimeter$

$X+X+\sqrt2 \cdot X=perimeter$

$2X+\sqrt2 X = perimeter$

And that's the solution!

According to the properties of the square, the diagonals intersect each other, therefore, BE is equal to CE