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Circumscribed quadrilaterals, circumscribed polygons (higher level courseware)
We have already seen it that every triangle has an inscribed circle. Is the same true for quadrilaterals too? Counterexamples may be easily found, as no concave quadrilateral nor (among convex quadrilaterals) non-square rectangle has an inscribed circle (Figure 82). Henceforth we are going to deal with those quadrilaterals which have inscribed circles.
DEFINITION: The quadrilaterals which have inscribed circles are called circumscribed quadrilaterals.
Figure 82
Let us take a circumscribed quadrilateral and let us draw its inscribed circle. (Figure 83) The tangent line segments drawn from an external point to the circle are of equal length, therefore when denoting the line segments of equal length in the figure with the same letters we can realise that
\latex{AB + CD = x + y + z + u and AD + BC = x + u + z + y,}
which implies that AB + CD = AD + BC.THEOREM: The sums of the length of the opposite sides of a circumscribed quadrilateral are equal.
Figure 83
\latex{ B }
\latex{ C }
\latex{ D }
\latex{ A }
\latex{ x }
\latex{ u }
\latex{ u }
\latex{ x }
\latex{ y }
\latex{ z }
\latex{ z }
\latex{ y }
It can be proven that the converse of the theorem is also true for convex quadrilaterals, i.e.
THEOREM: If the sums of the length of the opposite sides of a convex quadrilateral are equal, then it is a circumscribed quadrilateral.
So in the case of convex quadrilaterals the sums of the length of the opposite sides being equal is a necessary and sufficient condition for the quadrilateral being a circumscribed quadrilateral. The combination of the theorem and of the converse of the theorem of circumscribed quadrilaterals can also be rephrased in other words:
THEOREM: A convex quadrilateral is a circumscribed quadrilateral if and only if the sums of the length of its opposite sides are equal.
Example 1
Let us prove that every convex kite has an inscribed circle.
Solution
From the definition two and two adjacent sides of a kite are of equal length, therefore using the notations of figure 84:
AB + CD = AD + BC = a + b.
The converse of the theorem of circumscribed quadrilaterals implies the statement of the example.
Figure 84
\latex{ A }
\latex{ C }
\latex{ B }
\latex{ D }
\latex{ b }
\latex{ a }
\latex{ a }
\latex{ b }
Example 2
The bases of a right-angled trapezium are \latex{4\;cm} and \latex{10\;cm} long. How long are the two legs, if we know that the trapezium is a circumscribed quadrilateral?
Figure 85
\latex{ T }
\latex{ B }
\latex{ A }
\latex{ C }
\latex{ D }
\latex{ r }
\latex{ r }
Solution
The leg perpendicular to the bases of the trapezium is \latex{2r} long, where r is the radius of the inscribed circle. The length of the other leg measured in cm is \latex{14-2r} according to the theorem of circumscribed quadrilaterals
If the base point of the line segment from C perpendicular to AB is T (Figure 85), then applying the Pythagorean theorem for triangle CTB: (14 – 2r)2 = 62 + (2r)2. After squaring: 196 – 56r + 4r2 = 36 + 4r2
which implies \latex{56r=160}. It implies \latex{r=\frac{20}{7}\,cm} \latex{\approx 2.86\,cm}. So \latex{AD=2r=}\latex{r=\frac{40}{7}\,cm} \latex{\approx 5.72\,cm}; \latex{BC=14\;cm} \latex{-\frac{40}{7}\,cm} =\latex{\frac{58}{7}\,cm} \latex{\approx 8.29\,cm}.
DEFINITION: A polygon is called a circumscribed polygon if it has an inscribed circle.
Example 3
Let us prove that the sums of the length of the non-adjacent sides of a circumscribed polygon with even number of sides are equal, i.e. if A1, A2, …, A2n denote the consecutive vertices of the polygon according to an orientation, then
A1A2 + A3A4 + … + A2n –1A2n = A2A3 + A4A5 + … + A2nA1.
Figure 86
A1
A2
A3
A4
A5
A6
\latex{ a }
\latex{ b }
\latex{ b }
\latex{ c }
\latex{ c }
\latex{ d }
\latex{ d }
\latex{ e }
\latex{ e }
\latex{ f }
\latex{ f }
\latex{ a }
Solution
For simplicity we are going to show the fulfilment of the statement for a hexagon only; the steps of the proof are the same for polygons with more sides. The tangent line segments drawn from an external point to the circle are of equal length, thus according to figure 86
A1A2 + A3A4 + A5A6 = a + b + c + d + e + f = A2A3 + A4A5 + A6A1.
Exercises
{{exercise_number}}. Prove that if a parallelogram is a circumscribed quadrilateral, then it is a rhombus.
{{exercise_number}}. What can we say about the interior angle bisectors of a circumscribed quadrilateral? (Justify your statement.)
{{exercise_number}}. Construct a rhombus if the radius of the inscribed circle and the length of one of its diagonals are given.
{{exercise_number}}. Construct a right-angled circumscribed trapezium if the radius of the inscribed circle and the length of one of the bases are given.
{{exercise_number}}. Three sides of a circumscribed quadrilateral are \latex{3\,cm}, \latex{4\,cm} and \latex{6\,cm} long. Calculate the length of the fourth side in each case.