One way would be to draw a square and then draw the diagonal of the square. You now have two isosceles triangles.
Another way would be to draw two line segments that intersect at a point. Take a compass and put the point at the intersection and set some arbitrary length to draw an arc. Now draw the arc between the two line segments to "connect" them. Then use a straight edge to draw another line segment between the points where the arc cuts the line segments, thus creating a triangle. The compass has marked off equal lengths along the line segments, so those sides of the triangle are equal. An isosceles triangle has two equal sides. As an extra point, an equilateral triangle is isosceles, but it is a special case of an isosceles triangle.
The later construction method is pretty straight forward (less involved) than the first, but both are 100% correct. And both methods can be accomplished using the classic "compass and straight edge method" from the days of the ancients!
You cannot. An isosceles triangle cannot be scalene and a scalene triangle cannot be isosceles. So an isosceles scalene triangle cannot exist.
The altitude line is perpendicular to the base and bisects the apex of the isosceles triangle.
Yes either an isosceles triangle or an equilateral triangle
yes
Its like an isosceles triangle that has been cut in half parrllel to its base
Yes to both
yes it is possible to do that.
It's impossible.
Yes, it is possible to draw an isosceles acute triangle. An isosceles triangle has at least two sides of equal length, and an acute triangle has all three angles less than 90 degrees. By ensuring that the two equal sides are longer than the base and that the angles opposite the equal sides are acute, one can create an isosceles acute triangle.
An isosceles triangle has 3 sides 2 of which must be equal in length.
you draw it with 2 sides of the same length and 1 that isn't the same as the others
It certainly can be. To draw one, draw a very flat "V" , one with a very wide angle, then connect the top to make a triangle.