a In the following diagram HIJK is in a .
We will show that a formed by connecting the to the circle at H, I, J, and K is also a square.
To show that quadrilateral ABCD is also a square, we will first prove that each interior angle of it is a . Notice that each side of HIJK is a of the circle and they form consecutive . Therefore, we will begin by recalling .
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Theorem 12-6
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Within a circle or in congruent circles, congruent chords have congruent arcs.
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With this theorem we know that there are four consecutive congruent arcs.
HK ≅ KJ ≅ JI ≅ IH
Since the of an entire circle is 360^(∘), we can find the arc measure of each arc by dividing 360^(∘) by the number of arcs.
360 ÷ 4 = 90
From here we can find m∠ A by considering .
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Theorem 12-14
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The measure of an angle formed by two lines that intersect outside a circle is half the difference of the measures of the intercepted arcs.
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In this case m∠ A will be half the difference of mHK and mHIK. Since the measure of each consecutive arc is 90 ^(∘), the measure of mHIK is 270^(∘).
Now we are ready to find m∠ A.
m∠ A= 1/2(mHIK-mHK)
m∠ A= 1/2( 270- 90)
m∠ A= 1/2(180)
m∠ A=90
We have a . Next, we will show that all the sides of ABCD are congruent. To do so we will recall .
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Theorem 12-3
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If two tangent segments to a circle share a common endpoint outside the circle, then the two segments are congruent.
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By this theorem we can write the following congruence relations.
&AH ≅ AK BH ≅ BI
& CI ≅ CJ DJ ≅ DK
We can see that we have four .
Since they are also we can conclude that the base angles of each triangle are 45^(∘). Additionally, the bases of each triangle are congruent given that HIJK is a square.
Looking at the diagram, we can see that for each triangle the base angles and the side included between the base angles are congruent. Therefore, by the the triangles are congruent. Note that corresponding parts of congruent triangles are congruent.
Finally, by the definition of a square we can conclude that ABCD is a square.
