Let's draw the isosceles trapezoid ABCD, where AD ≅ BC, and mADC ≅ mBCD.
If we draw the diagonals AC and BD of the trapezoid two congruent triangles are formed,
∆ ADC ≅ ∆ BDC (SAS Postulate: If two sides and the angle between them in one triangle are congruent to the corresponding parts in another triangle, then the triangles are congruent).
Since these triangles are congruent, AC ≅ BD.
prove any two adjacent triangles as congruent
Suppose that the parallelogram is a rhombus (a parallelogram with equal sides). If we draw the diagonals, isosceles triangles are formed (where the median is also an angle bisector and perpendicular to the base). Since the diagonals of a parallelogram bisect each other, and the diagonals don't bisect the vertex angles where they are drawn, then the parallelogram is not a rhombus.
A ruler or a compass would help or aternatively use Pythagoras' theorem to prove that the diagonals are of equal lengths
You can't. A trapezoid is a quadrilateral because it has four sides. The definition of a quadrilateral is a shape with four sides.
it would produce two right angle triangleImproved Answer:-Measure them and use a protractor which will result in equal measures of 4 by 90 degrees angles
You can't because triangles do not have diagonals but an isosceles triangle has 2 equal sides
You prove that the two sides (not the bases) are equal in length. Or that the base angles are equal measure.
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.
To prove a trapezoid is isosceles, you need to show that the legs (the non-parallel sides) are congruent. This can be done by demonstrating that the base angles opposite these sides are congruent. You can use the triangle congruence postulates or the properties of parallel lines and transversals to establish the equality of these angles.
prove any two adjacent triangles as congruent
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.
Check to see that -- it has four sides -- its two diagonals are equal.
Suppose that the parallelogram is a rhombus (a parallelogram with equal sides). If we draw the diagonals, isosceles triangles are formed (where the median is also an angle bisector and perpendicular to the base). Since the diagonals of a parallelogram bisect each other, and the diagonals don't bisect the vertex angles where they are drawn, then the parallelogram is not a rhombus.
A ruler or a compass would help or aternatively use Pythagoras' theorem to prove that the diagonals are of equal lengths
Because its base angles are equal and the 4 interior angles of any quadrilateral add up to 360 degrees so in this case opposite angles must add up to 180 degrees
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
It has 3 sides 2 of which are equal in length It has 3 angles 2 of which are the same size