The Trapezoid Midsegment Theorem states that a line segment connecting the midpoints of the legs of the trapezoid is parallel to the bases, and equal to half their sum. Here, we'll prove the … [Read more...] about Converse of the Trapezoid Midsegment Theorem
Quadrangles (also called quadrilaterals) are polygons with 4 edges, and 4 angles, or vertices, at the corners.
There are several types of special polygons, with unique properties that depend on factors such as:
1. If their sides are parallel to each other or not - and if they are parallel, are both pairs of opposite edges parallel, or just one set
2. If their sides are all equal or if they just have two pairs of equal sides.
3. If the angles are right angles.
We have a section for each special type of polygon, describing and proving their properties, which are very common in high school geometry problems.
The sum of the angles in a quadrangle
One property that is common to all quadrangles, other than 4 sides and 4 vertices, is that the sum of the angles in a quadrangle is always 360°. The proof of this is simple.
Proof: Show that the sum of interior angles in a simple convex quadrangle is always 360°
Let’s think about the strategy to do this. 360 is a familiar number - it is the measure of angles in a circle, but there are no circles here, so let’s rule that out.
360 is also 2x180, and that is also a familiar number – it is the sum of the angles in a triangle – so, if we can show that this quadrangle is composed of two triangles we will have the proof- and that’s all we need.
Proof: In any simple convex polygon, a line connecting 2 points on the perimeter of the polygon is entirely within the polygon, per the definition of convex polygons. So, let’s draw a line connecting two opposite corners of the quadrangle (such a line is a called a ‘diagonal’) – say from C to A:
(1) m∠A1 + m∠D+ m∠C1 = 180° //sum of the interior angles in triangle ΔADC
(2) m∠A2 + m∠B+ m∠C2 = 180° //sum of the interior angles in triangle ΔABC
(3) m∠A1 + m∠D+ m∠C1+m∠A2 + m∠B+ m∠C2 = 360° //add both equations
(4) m∠A1 + m∠A2 +m∠D+ m∠C1+ m∠C2 + m∠B= 360° //re-arrange terms
(5) m∠A1 + m∠A2 = m∠A // angle addition postulate
(6) m∠C1 + m∠C2 = m∠C // angle addition postulate
(7) m∠A+m∠D+ m∠C + m∠B= 360° ∎ //substitution property of equality
Now that we've explained the basic concept of quadrangles in geometry, let's scroll down to work on specific geometry problems relating to this topic.
A quadrilateral whose diagonals bisect each other is a parallelogram, as we will show in this exercise. One of the properties of a parallelogram is that its diagonals bisect each other. This is a … [Read more...] about Quadrilateral Whose Diagonals Bisect Each Other
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A rhombus is a special kind of parallelogram, in which all the sides are equal. We've seen that one of the properties of a rhombus is that its diagonals are perpendicular to each other. Here we will … [Read more...] about A Parallelogram with Perpendicular Diagonals is a Rhombus
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In today's lesson, we will prove that the interacting diagonals of a parallelogram divide it into four triangles, all of which have equal areas. Each of the diagonals of a parallelogram divides it … [Read more...] about The Diagonals Divide a Parallelogram Into Four Equal Areas
In a square, the diagonals bisect each other. This is a general property of any parallelogram. And as a square is a special parallelogram, which has all the parallelogram's basic properties, this is … [Read more...] about The Diagonals of a Square Bisect Each Other
We have shown that in any parallelogram, the opposite angles are congruent. Since a rhombus is a special kind of parallelogram, it follows that one of its properties is that both pairs of opposite … [Read more...] about Properties of Rhombus: The Opposite Angles are Congruent
A square is a geometric shape which is fully determined by the lengths of its side, a. If we know the length of the side of a square, we know its perimeter, its area, the length of its diagonals, … [Read more...] about How to Find the Diagonal of a Square