The theorems here are ones I think are important to the overall development of Euclidean Geometry but may not be featured or covered in the text. We will have discussion in class about pursuing these theorems. Some of them may be embedded in the Problem Sets.
Given a circle and two chords AB and CD intersecting at point P in the circle, then the products of the parts of each chord are equal. That is:
(AP)(PB) = (CP)(PD)
Given a circle and a tangent segment PT and secant AB from the same external point P, the square of the tangent segment is equal to the product of the secant segment PA and its external portion PB. That is:
Given a circle and two secants from an external point P, the product of the measures on one secant segment and its external portion is equal to the product of the measures of the other segment segment and its external portion. That is:
(PA)(PB) = (PC)(PD)
SupThm4. Ptolemy's Theorem
In a cyclic quadrilateral, the product of the diagonals is equal to the sum of the products of pairs of opposite sides. That is,
(BD)(AC) = (AD)(BC) + (AB)(CD)
SupThm5. Converse of SupThm4.
Given a quadrilateral ABCD with the sum of the products of the two pairs of opposite sides equal to the product of the two diagonals. That is, AD·BC + AB·CD = AC·BD. Prove that ABCD is inscribed in a circle.
SupThm6. Ceva's Theorem
Given a triangle ABC with an arbitrary point P inside the triangle. Any segment from a vertex to the intersection with the opposite side is called a Cevian. Let D, E, and F be the feet of the Cevians opposite A, B,and C respectively.
If the three Cevians AD, BE, and CF are concurrent at P, then prove
then the Cevians AD, BE, and CF are concurrent.
SupThm7. Theorem of Menelaus
Given any line that transverses (crosses) the three sides of a triangle (one of them will have to be extended), six segments are cut off on the sides. The product of three non-adjacent segments is equal to the product of the other three. The converse also holds.
Usually, the two products are expressed in a ratio equal to 1 or equal to -1 if directed segments are used.
The Theorem of Menelaus is perhaps a misnomer. Menelaus produced an analogous theorem for spherical geometry. The theorem for the geometry of the plane was known before Menelaus. Most geometry references, however, cite this theorem for plane geometry as Menelaus's Theorem.
The bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides. That is, for any triangle ABC, the bisector of the angle at C divides the opposite side into segments of length AD and DB such that
If triangle ABC is not isosceles, prove that the bisector of an exterior angle of a triangle divides the opposite side externally into segments that are proportional to the adjacent sides.
That is, the external bisector of the angle at C externally divides the side AB at M such that
Why do we need to exclude the case when triangle ABC is isosceles. What is the ratio of the adjacent sides when the triangle is isosceles?
Recall that in a triangle, the internal bisector and the external bisector of an angle are perpendicular. Therefore if CD and CM are drawn in the same triangle, they will meet in a right angle at C.
Lemma 1. Given a set of triangles all having the same base AB. What is the locus of the vertex C in the ratio of the sides adjacent to C is 1? Proof?
Lemma 2. Given a set of triangles all having the same base AB. What is the locus of the vertex C in the ratio of the sides adjacent to C is not equal to 1? Proof? Build a GSP animation showing this locus.
SupThm 9 Law of Sines
Prove this version of the Law of Sines by using the inscribed angle theorem.
Through the midpoint M of any chord PQ of a circle, any chords AB and CD are drawn; chords AD and BC meet PQ at points X and Y. Prove that M is the midpoint of XY. Constructing a valid picture of this situation is a part of the problem, but only the first step in building the proof.