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An isosceles triangle has 3 sides 2 of which are equal in length An isosceles triangle has 3 interior angles 2 of which are the same size
an isisceles triangle is a triangle with atleast two sides congruent.
You can't because triangles do not have diagonals but an isosceles triangle has 2 equal sides
If You Prove An Isosceles Triangle To Have Three Equal Sides. You Now Have Disproved It As Being An Isosceles Triangle. So Even If You Could You Would Now Have An Equilateral Triangle. I Just Can`t See A Way This Can Be Done.
If you can only prove two sides of an apparently equilateral triangle to be congruent then you have to use isosceles.
What have we got to prove? Whether we have to prove a triangle as an Isoseles triangle or prove a property of an isoseles triangle. Hey, do u go to ALHS, i had that same problem on my test today. Greenehornet15@yahoo.com
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It has 3 sides 2 of which are equal in length It has 3 angles 2 of which are the same size
its half a square and a square is 360
Yes it is possible. Obtuse means the triangle contains an angle which is greater than 90 degrees, and isosceles means the triangle has two sides of the same length. So to prove this in the easiest way possible, you can make a dot on your page, measure 91 or more degrees and draw two equal length lines out at this angle, then connect these two lines to make an obtuse isosceles triangle.
Suppose the diagonals meet at a point X.AB is parallel to DC and BD intersects themTherefore, angle ABD ( = ABX) = BAC (= BAX)Therefore, in triangle ABX, the angles at the ends of AB are equal => the triangle is isosceles and so AX = BX.AB is parallel to DC and AC intersects themTherefore, angle ACD ( = XCD) = BDC (= XDC)Therefore, in triangle CDX, the angles at the ends of CD are equal => the triangle is isosceles and so CX = DX.Therefore AX + CX = BX + DX or, AC = BD.
The theorem is only true if the base is the side of different length.To see this consider the right angled isosceles triangle with sides 1, 1 and √2. If one of the sides of length 1 is the base, the height is obviously the other side of length 1, but it clearly does not meet the base at its mid-point to make it a median.So with an isosceles triangle ABC with sides AB & AC equal, angles ABC & ACB equal and side BC the base, we need to prove that the point X where the height (AX) meets BC is such that BX = CX.Consider triangles AXB and AXC.Angle AXB is a right angle, as is AXC (since AX is a height of triangle ABC).Side AB is the hypotenuse of triangle AXB; AC is the hypotenuse of triangle AXC; they are known to be equal (from isosceles triangle ABC)Side AX is common to both trianglesThus triangles AXB and AXC are congruent since we have a Right-angle, Hypotenuse, Side match.Thus XB must be the same length as XC, that is X is the mid-point of BC.As X is the mid-point of BC, AX is the median.