 # Geometry for College Students

Let P,Q, and R be the points of tangency of the incircle of ∆ABC with sides BC, CA, and AB, respectively. Then lines AP, BQ, and CR:
are concurrent.
Gergonne point
Point of congruency of lines AP, BQ, and CR, where P,Q, and R are the points of tangency of the incircle of ∆ABC.
Ceva’s Theorem:
Let AP, BQ, and CR be three lines joining the vertices of ∆ABC to pints P, Q, and R on the opposite sides. Then these three lines are concurrent if and only if:
(AR/RB)(BP/PC)(CQ/QA) = 1
Given distinct points A and B and a positive number µ, there is exactly ______ point X such that _________________. Also there is at most _________ other point on the line ____ for which this equation holds.
one
AX/XB = µ
one
AB
Let AP, BQ, and CR be Cevians in ∆ABC, where P, Q, and R lie on sides BC, CA, and AB respectively. If these Cevians are concurrent at point T, then:
AT/AP = [(AR·CQ) + (QA·RB)] / [AR·CQ) + (QA·RB) + (RB·CQ)]
Let T be the Gergonne point of ∆ABC. If coincides with the incenter or the circumcenter or the orthocenter or the centroid of ∆ABC, then:
The triangle must be an equilateral.
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Let U and V be points on sides of AB and AC, respectively, of ∆ABC and suppose UV is parallel to BC. Then the intersection of UC and VB:
lies on the median of AM.
The lines joining the vertices of a triangle to the points of tangency of the opposite exscribed circles are:
concurrent.
Given three concurrent Cevians in a triangle, the the lines obtained by joining the midpoints of the Cevians to the midpoints of the coorresponding sides are:
concurrent.
Suppose Cevians AP, BQ, and CR are concurrent at point T so that ∆ABC is decomposed into six small triangles. If areas of ∆ART, ∆BPT, and ∆CQT are equal, then:
All six small triangles have equal areas.
General Form of Ceva’s Theorem:
Let AP, BQ, and CR be Cevians of ∆ABC, where points P, Q, and R lie on lines BC, CA, and AB, respectively. Then these Cevians are either concurrent OR parallel if and only if:
an ODD number of them are interior and (AR/RB)(BP/PC)(CQ/QA) = 1
In figure 4.5 on p 135, BQ/QC = ___
and XP/PY = ______. And if XP/PY =BQ/QC, then:
st/ru; [s(t+u)]/[u(r+s)]
then both ratios are equal to 1 and in that case XY is parallel to BC.
If Cevians AP, BQ, and CR are parallel, then:
the Cevian product is trivial.
Angular Cevian Product:
Suppose that AP, BQ, and CR are Ceviansin ∆ABC. Then the corresponding Cevian product is equal to:
[sin(∠ACR)sin(∠BAP)sin(∠CBQ)]/[sin(∠RCB)sin(∠PAC)sin(∠QBA)]
In particular, the three Cevians are concurrent or parallel if and only if an odd number of them are interior and this angular Cevian product is equal to 1.
Given an arbitrary ∆ABC, build three outward-pointing triangles BCU, CAB, and ABW, each sharing a sides with the original triangle. Assume that ∠BAW = ∠CAV, ∠CBU = ∠ABW, and ∠ACV = ∠BCU. And assume that AU, BV, and CW cut across the interior of ∆ABC. Then:
lines AU, BV, and CW are concurrent.
The pedal triangle of acute ∆ABC is ∆DEF, and perpendicular AU, BV, and CW are dropped from the vertices of the original triangle to the sides of the pedal triangle. Then lines AU, BV, and CW are:
concurrent and the point of concurrence is at the circumcenter.
If Cevians AP, BQ, and CR of ∆ABC are concurrent or parallel, then:
their isogonal Cevians are also concurrent or parallel.
Isogonal line:
The line reflected across an angle bisector.
Symmedians:
The isogonals of the medians.
Lemoine point:
The point of concurrence of the symmedians.
Isogonal conjugate:
The point of concurrence of the isogonal lines.
Let X be a point other than A, B, or C on the circumcircle of ∆ABC. Then:
the isogonal Cevians of AX, BX, and CX are parallel.
Let X be any point in the interior of ∆ABC and let Y be the circumcenter of the triangle whose vertices are the reflections of X in the sides of ∆ABC. Then:
Y is the isogonal conjugate of X with respect to ∆ABC.
Let ABCDEF be a hexagon inscribed in a circle. Then the diagonals AD, BE, and CF are concurrent if and only if:
(AB/BC)(CD/DE)(EF/FA)=1
Given and acute angled ∆ABC, show that the lines joining A, B, and C to the midpoints of the nearer sides of the pedal triangle are:
concurrent
The only point in the interior of ∆ABC that is its own isogonal conjugate is:
the incenter.
If ∆ABC is not a right triangle then the circumcenter and orthocenter are:
isogonal conjugates.
If ∠C = 90 in ∆ABC, then the Lemoine point of the triangle lies on:
the altitude from vertex C.
If the Lemoine point of ∆ABC lies on the altitude from vertex C, then:
either AC = BC or ∠C = 90.