Given:

Square ABCD
Red circle c1
Green circle c2
Points H and S are tangent points for c1 and ABCD
J is the point of tangency for c1 and c2
J lies on the diagonal of ABCD

Prove: Points H, S, and J form an isosceles triangle.

To start this proof, let's first construct the segments from points H and S to the center point of the red circle, point I. If we can show that triangles HJI and SJI are congruent, then by the Congruent Parts of Congruent Figures (CPCF) Theorem, then we will know that segments HJ and JS are congruent. Therefore by the definition of and isosceles triangle, triangle HJS is isosceles.

By the definition of a circle, segments HI and IS are congruent and by the reflexive property, we know that segment IJ is congruent to itself. All we need to show is that a(HIJ) = a(SIJ) and the triangles are congruent by the Side-Angle-Side (SAS) Congruence Theorem.

By definition of tangency, we know that a(IHA) = a(ISA) = 90 degrees. We already know that a(A) is a right angle (therefore = 90 degrees) from the given square. Therefore, AHIS is a square as well. Since AHIS is a square we also know that diagonal AI bisects angle HIS. Therefore, a(HIA) = a(SIA). Recall from the given that center point I (from the red circle) and tangent point J is on diagonal AD. Therefore, by the definition of supplementary angles,

Since we've already established that a(HIA) = a(SIA), then by substitution and simplification we also know that a(HIJ) = a(SIJ). Since this is true, we can now say, by the SAS Theorem, that triangles HJI and SJI are congruent. Therefore, d(HJ) = d(SJ), which means that triangle HJS is isosceles by definition.

To see other proofs concerning this particular construction, please continue onto Page 2.

Continue to the Conclusion Page

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