Category Archives: Geometry

CGMO 2012 ( China Girls Math Olympiad 2012) Problem 5

As shown in the figure below, the in-circle of ABC is tangent to sides AB and AC at D and E respectively, and O is the circumcenter of BCI. Prove that \angle ODB = \angle OEC.
import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xma...

CGMO 2012 (China Girls Math Olympiad 2012)- Problem 2

Circles Q_1 and Q_2 are tangent to each other externally at T. Points A and E are on Q_1, lines AB and DE are tangent to Q_2 at B and D, respectively, lines AE and BD meet at point P
Prove that

(1) \frac{AB}{AT}=\frac{ED}{ET};
(2) \angle ATP + \angle ETP = 180^{\circ}.
import graph; size(5.97cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real ...

Important theorems of Geometry(Triangles) – 1

Today we will be discussing about  2 important Triangle theorems which have very High applications in field of Geometry

1) Angle Bisector theorem
2) Stewart’s Theorem

Angle bisector theorem :

Introduction :

The Angle Bisector Theorem states that given triangle \triangle ABC and angle bisector AD, where D is on side BC, then c/m= b/n . Likewise, the converse of this theorem holds as well.

size(200);defaultpen(fontsize(10));real a,b,c,d;pair A=(1,4), B=(-5,0), C=(3,0), D;b = abs(C-A);c = abs(B-A);D = (b*B+c*C)/(b...

Proof :

Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick:

size(200);defaultpen(fontsize(10));real a,b,c,d,m,n;pair A=(1,4), B=(-5,0), C=(3,0), D, E;b = abs(C-A);c = abs(B-A);D = (b*B+...

Since AB and CE are parallel, we know that \angle BAE=\angle CEA and \angle BCE=\angle ABC. Triangle ACE is isosceles, with AC = CE.

By AA similarity\triangle DAB \cong \triangle DEC. By the properties of similar triangles, we arrive at our desired result:

c/m = b/n

Stewarts Theorem :

Introduction :

Given a triangle \triangle ABC with sides of length a, b, c opposite vertices ABC, respectively. If cevian AD is drawn so that BD = mDC = n and AD = d, we have that b^2m + c^2n = amn + d^2a. (This is also often written man + dad = bmb + cnc, a form which invites mnemonic memorization, e.g. “A man and his dad put a bomb in the sink.”)

Stewart's theorem.png

Proof :

Applying the Law of Cosines in triangle \triangle ABD at angle \angle ADB and in triangle \triangle ACD at angle \angle CDA, we get the equations

  • n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2}
  • m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2}

Because angles \angle ADB and \angle CDA are supplementarym\angle ADB = 180^\circ - m\angle CDA. We can therefore solve both equations for the cosine term. Using the trigonometric identity  \cos{\theta} = -\cos{(180^\circ - \theta)} gives us

  • \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}
  • \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n. However, m+n = a som^2n + n^2m = (m + n)mn and we can rewrite this as man + dad= bmb + cnc (A man and his dad put a bomb in the sink).
Thank you

Source :
Art of problem solving