Question Statement

A quadrilateral $ABCD$ is circumscribing a circle. Prove that: $|\overline{AB}|+|\overline{CD}|=|\overline{AD}|+|\overline{BC}|$

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Background and Explanation

This proof is based on the geometric property that tangent segments drawn from a common external point to a circle are equal in length. By applying this rule to each vertex of the quadrilateral, we can express the side lengths as sums of these equal segments.

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Solution

To prove the statement, we identify the points where the circle is tangent to the sides of the quadrilateral. Let the points of tangency on sides $AB,BC,CD,$ and $DA$ be $E,F,G,$ and $H$ respectively.

As we know that the tangents drawn to a circle from a point lying outside the circle are equal in length, we have the following equalities:

\begin{align*} |\overline{AE}| &= |\overline{AH}| \\ |\overline{BE}| &= |\overline{BF}| \\ |\overline{CG}| &= |\overline{CF}| \\ |\overline{DG}| &= |\overline{DH}| \end{align*}

Step 1: Summing segments for side $AB$

Adding equations (i) and (ii): $\begin{array}{rl}|\overline{AE}|+|\overline{BE}|& =|\overline{AH}|+|\overline{BF}|\\ \Rightarrow |\overline{AB}|& =|\overline{AH}|+|\overline{BF}|\end{array}$ (Since $E$ lies on $AB$, $|\overline{AE}|+|\overline{BE}|=|\overline{AB}|$)

Step 2: Summing segments for side $CD$

Adding equations (iii) and (iv): $\begin{array}{rl}|\overline{CG}|+|\overline{DG}|& =|\overline{CF}|+|\overline{DH}|\\ \Rightarrow |\overline{CD}|& =|\overline{CF}|+|\overline{DH}|\end{array}$ (Since $G$ lies on $CD$, $|\overline{CG}|+|\overline{DG}|=|\overline{CD}|$)

Step 3: Final Addition

Now, we add equations (v) and (vi) together to find the sum of the opposite sides $AB$ and $CD$: $\begin{array}{rl}|\overline{AB}|+|\overline{CD}|& =|\overline{AH}|+|\overline{BF}|+|\overline{CF}|+|\overline{DH}|\end{array}$

By rearranging the terms on the right-hand side to group segments that lie on the same side of the quadrilateral: $\begin{array}{rl}|\overline{AB}|+|\overline{CD}|& =(|\overline{AH}|+|\overline{DH}|)+(|\overline{BF}|+|\overline{CF}|)\end{array}$

Since $|\overline{AH}|+|\overline{DH}|=|\overline{AD}|$ and $|\overline{BF}|+|\overline{CF}|=|\overline{BC}|$, we get: $|\overline{AB}|+|\overline{CD}|=|\overline{AD}|+|\overline{BC}|$

This proves that the sum of the lengths of opposite sides of a circumscribed quadrilateral are equal.

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Key Formulas or Methods Used

• Tangent Segment Theorem: Tangents drawn from an external point to a circle are equal ($PA=PB$).
• Segment Addition Postulate: If a point $B$ lies on a line segment $AC$, then $AB+BC=AC$.

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Summary of Steps

1. Identify the four points of tangency on the sides of the quadrilateral.
2. Set up four equations based on the property that tangents from the same vertex are equal.
3. Add the equations corresponding to the segments of side $AB$ and side $CD$.
4. Combine the resulting segments on the right side of the equation to form the lengths of sides $AD$ and $BC$.