Geometry is perhaps the most important topic in mathematics as far as Math Olympiad and I.S.I. Entrance goes. The following list of results may work as an elementary set of tools for handling some geometry problems.

‘Learning’ them won’t do any good. One should ‘find’ these ‘truths’ through intense problem solving and active imagination.

The following is a suggestive list (derived from existing texts).

**Warning: Full statements are NOT given here. It is supposed to be used as a compliment to Cheenta Classroom Program. **

**Mid-Point Theorem:**Segment joining midpoints of two sides of a triangle are parallel to and half of the third side. The converse is true.**Parallel lines:**Parallel lines split sides proportionally.**Area-altitude -base:**If one of the three quantities are fixed, the other two are proportional.- Similar triangles have the ratio of areas = ratio of squares on corresponding sides.

**Equiangular = Similar = Sides proportional****Similar Triangle Criterion:**Ratio of pair of sides equivalent**and**included angle same.**Angle Bisector Theorem:**Angle bisector divides the third side into the ratio of containing sides. External bisector does something similar.**Trapezium Bifurcation:**Intersect the diagonals at P, intersect the non-parallel sides at Q. Then PQ bisects the parallel sides.**Angles in the circle:**Angle at the center is twice the angle at the circumference (subtended by the same arc).**Hence**angles subtended by the same arc at the circumference are equal. This also works as a**condition for****concyclic points.****Cyclic Quadrilateral:**Opposite points add up to \( 180^o\). This also works as a**condition for****concyclic points.****Tangent – Chord Theorem:**Angle subtended by a chord in the opposite arc, is equal to the angle made with the tangent at an end of the chord.**Power of point:**Fix a circle G. Fix a point P inside, on or outside the circle. Suppose PAA’ and PBB’ are two chords or secants from P to G. Also if possible draw PT, the tangent. The power of the point P is $$ PA \times PA’ = PB \times PB’ = PT^2 $$ In particular those quantities are equal.**Ceva’s Theorem:**Concurrent cevians divide opposite sides into ratios which when suitably multiplied gives one. The**converse is a condition of concurrency**. Trigonometric form exists.**Menelaus Theorem:**Collinear points split sides into pieces whose ratios multiply to 1. The converse is a**condition for collinearity**.**Pythagoras Theorem:**\( \textrm{perpendicular}^2 + \textrm{base}^2 = \textrm{hypotenuse}^2 \). Special modifications give inequalities for acute and obtuse angled triangles.**Concurrent Lines in a Triangle:**Medians, altitudes, angle bisectors of a triangle are respectively concurrent at the centroid, orthocenter, and incenter.**Euler Line:**O-G-H are collinear. GH= 2OG.**Orthocenter distance:**AH, the altitude meets sides BC of a triangle at P and the circumference at D. Then DP= PH.**In-radius and Ex-radius:**\( \Delta = rs = r_1 (s-a) = r_2 (s-b) = r_3(s-c) \). Hence $$ \frac {1}{r} = \frac{1}{r_1} +\frac{1}{r_2} + \frac{1}{r_3} $$**Nine Point Circle:**Feet of altitudes, midpoints of sides of a triangle are collinear (with three other special points). Nine point center is at the mid point of OH.- Angle Bisector of \( \angle BAC \) is equal to the angle bisector of \( \angle OAH \)
**Euler Identity:**\( OI^2 = R^2 – 2Rr \)**Simson Line:**Feet of perpendiculars dropped from any point on the circumference to the sides of the triangle are collinear.**Ptolemy’s Theorem:**If ABCD is any quadrilateral then \( BC \times AD + AB \times CD \ge AD \times BC \). Equality holds for cyclic quadrilaterals.**Brahmagupta’s Theorem:**In \( \Delta ABC \), AD and AE be altitude and diameter. Then \( AB \times AC = AD \times AE \).