374. Some proofs of the Law of Cosines.

by Virgil Nicula, May 12, 2013, 2:59 PM

PP1. Prove that in any $\triangle ABC$ there is the relation $a^2=b^2+c^2-2bc\cos A$ , i.e. $\boxed{\cos A=\frac {b^2+c^2-a^2}{2bc}}$ - Law of Cosines.

Proof 1. Denote $D\in AC\ ,\ BD\perp AC$ . Thus, $\left\|\begin{array}{c} DB=c\sin A\\\\ DC=b-c\cos A\end{array}\right\|$ . Here appear two cases, $D\in (AC)$ or $A\in (CD)$ . Apply the Pythagoras' theorem

in a $D$-right-angled $\triangle BDC\ :$ $BC^2=$ $DB^2+DC^2\iff$ $a^2=c^2\sin^2A+\left(b-c\cos A\right)^2\iff$ $a^2=b^2+c^2-2bc\cos A\iff$ $\boxed{\cos A=\frac {b^2+c^2-a^2}{2bc}}$ .

Proof 2. Let $D\in BC\ ,\ AD\perp BC$ . Suppose w.l.o.g. that $D\not\in\{B,C\}$ $\implies$ $\left\|\begin{array}{ccccc} \mathrm{Case}\ 1\ :\ D\in (BC) & \implies & \left|\begin{array}{c} DB=c\cos B\\\\ DC=b\cos C\end{array}\right| & \implies & BC=DB+DC\\\\ \mathrm{Case}\ 2\ :\ B\in (DC) & \implies & \left|\begin{array}{c} DB=-c\cos B\\\\ DC=b\cos C\end{array}\right| & \implies & BC=DC-DB\\\\ \mathrm{Case}\ 3\ :\ C\in (BD) & \implies & \left|\begin{array}{c} DB=c\cos B\\\\ DC=-b\cos C\end{array}\right| & \implies & BC=DB-DC\end{array}\right\|$ $\implies$

$\boxed{b\cos C+c\cos B=a}$ . Hence $\left\{\begin{array}{ccc} \underline{b\cos C}+\underline{\underline{c\cos B}}=a & \odot & (-a)\\\\ c\cos A+\underline{a\cos C}=b & \odot & b\\\\ \underline{\underline{a\cos B}}+b\cos A=c & \odot & c\end{array}\right|\bigoplus\implies$ $a^2=b^2+c^2-2bc\cos A\iff \boxed{\cos A=\frac {b^2+c^2-a^2}{2bc}}$ .

Proof 3. Denote the circle $w=\mathbb C(B,c)$ , its diameters $[AG]$ , $[DE]$ , where $B\in (CE)$ and $\{A,F\}=\{AC\}\cap w$ .

Suppose w.l.o.g. $A\ne 90^{\circ}$ . Power of $C$ w.r.t. $w\implies\boxed{CD\cdot CE=CA\cdot CF}\ (*)\implies$

$\left\|\begin{array}{cccc} \mathrm{Case}\ 1\ :\ A<90^{\circ}\ \implies & \left|\begin{array}{c} AF=2c\cos A\\\\ CD=c-a\end{array}\right| & \stackrel{(*)}{\implies} & (c-a)(c+a)=b\left(2c\cos A-b\right)\\\\ \mathrm{Case}\ 2\ :\ A>90^{\circ}\ \implies & \left|\begin{array}{c} AF=-2c\cos A\\\\ CD=a-c\cos C\end{array}\right| & \stackrel{(*)}{\implies} & (a-c)(c+a)=b\left(b-2c\cos A\right)\end{array}\right\|$ $\implies a^2=b^2+c^2-2bc\cos A$ $\implies \boxed{\cos A=\frac {b^2+c^2-a^2}{2bc}}$ .

Remark 1. $\boxed{4S=\left(b^2+c^2-a^2\right)\tan A}$ . Indeed, $S=\frac {bc\sin A}{2}=$ $\frac {\left(2bc\cos A\right)\tan A}{4}=\frac {\left(b^2+c^2-a^2\right)\tan A}{4}\iff$ $4S=\left(b^2+c^2-a^2\right)\tan A$ .

Remark 2. Let $w=\mathbb C(O,R)$ and its exterior $C$ , i.e. $OC>R$ . Denote $\{A,B,M,N\}\subset w$ so that $AB$ separates $M$ , $N$ and $CA$ , $CB$ are tangent to $w$ .

In this case $AB=2R\sin\widehat{AMB}=$ $2R\sin\widehat{ANB}=2R\sin \widehat{CAB}$ - Law of Sines. Indeed, if $[AD]$ is diameter of $w$ , then $\sin \widehat {ADB}=\sin\widehat{AMB}$ and in

$D$-right $\triangle ADB$ we have $AB=AD\sin\widehat{ADB}\implies AB=2R\sin\widehat {AMB}$ .

Remark 3. Prove easily that $:\ \triangle ABC$ is acute $\iff S=[ABC]=[BOC]+[COA]+[AOB]\iff$ $2S=R\sum a\cos A=$ $\iff$ $\boxed{a\cos A+b\cos B+c\cos C=\frac {2S}{R}}\ ;$

$\triangle ABC$ is acute $\iff S=[ABC]=[BOC]+[COA]-[AOB]\iff$ $2S=R\cdot\left(b\cos B+c\cos C-a\cos A\right)=$ $\iff$ $\boxed{b\cos B+c\cos C-a\cos A=\frac {2S}{R}}\ .$

$\left\{\begin{array}{cc} b\cos C+c\cos B=a\\\\ c\cos A+a\cos C=b\\\\ a\cos B+b\cos A=c\\\\ a\cos A+b\cos B+c\cos C=\frac {2S}{R}\end{array}\right|\ \bigoplus\ \implies\ (a$ $+b+c)(\cos A+\cos B+\cos C)=(a+b+c)+\frac {r(a+b+c)}{R}\implies$ $\boxed{\cos A+\cos B+\cos C=1+\frac rR}\ .$

Remark 4. Let $D$ be the symmetrical point of $A$ w.r.t. the midpoint of the side $[BC]$ . Apply the Euler's relation in the quadrilateral (parallelogram) $ABDC\ :$

$4m_a^2+a^2=$ $2\left(b^2+c^2\right)$ . From the sum of the relations $\left\{\begin{array}{ccc} b^2+c^2-a^2 & = & 2bc\cos A\\\\ 4m_a^2+a^2 & = & 2\left(b^2+c^2\right)\end{array}\right\|\bigoplus\implies$ $\boxed{b^2+c^2+2bc\cos A=4m_a^2}$ .

Remark 5. $\left\|\begin{array}{ccccc} s(s-a) & + & (s-b)(s-c) & = & bc\\\\ s(s-a) & - & (s-b)(s-c) & = & bc\cos A\end{array}\right\|\implies$ $\left\|\begin{array}{ccccc} s(s-a) & = & \frac 12\cdot bc(1+\cos A) & = & bc\cos^2\frac A2\\\\ (s-b)(s-c) & = & \frac 12\cdot bc(1-\cos A) & = & bc\sin^2\frac A2\end{array}\right\|\implies$ $\left\|\begin{array}{ccc} \cos\frac A2 & = & \sqrt {\frac {s(s-a)}{bc}}\\\\ \sin\frac A2 &= & \sqrt {\frac {(s-b)(s-c)}{bc}}\end{array}\right\|$ .


PP2 (Ptolemy thorem). Let $ABCD$ be a convex and cyclic quadrilateral. Then there is the relation $AB\cdot CD+AD\cdot BC=AC\cdot BD$ .

Proof (trigonometric). Let the circumcircle $w=\mathbb C(O,R)$ of $\triangle ABC\ ,\ \left\{\begin{array}{c} AB=a\ ;\ BC=b\ ;\ CD=c\\\\ DA=d\ ;\ AC=e\ ;\ BD=f\end{array}\right\|$ and $\left\|\begin{array}{ccc} m\left(\widehat{ACB}\right)=m\left(\widehat{ADB}\right)=\alpha & \implies & a=2R\sin\alpha\\\\ m\left(\widehat{BAC}\right)=m\left(\widehat{BDC}\right)=\beta & \implies & b=2R\sin\beta\\\\ m\left(\widehat{CAD}\right)=m\left(\widehat{CBD}\right)=\gamma & \implies & c=2R\sin\gamma\\\\ m\left(\widehat{DBA}\right)=m\left(\widehat{DCA}\right)=\delta & \implies & d=2R\sin\delta\end{array}\right\|$ ,

where $\alpha+\beta+\gamma +\delta =180^{\circ}$ . In conclusion, $AB\cdot CD+AD\cdot BC=AC\cdot BD\iff$ $ac+bd=ef\iff$ $\boxed{\sin\alpha\sin\gamma+\sin\beta\sin\delta=\sin (\alpha +\beta)\sin (\alpha +\delta )}\iff$

$\cos (\alpha -\gamma )-\cos (\alpha +\gamma )+\cos (\beta -\delta )-\cos (\beta +\delta )=\cos (\beta -\delta )+\cos (\alpha -\gamma )\iff$ $\cos (\alpha +\gamma )+\cos (\beta +\delta )=0$ , what is truly because $(\alpha +\gamma )+(\beta +\delta )=180^{\circ}$ .

Particular case. If $[AC]$ is diameter, i.e. $AC=2R$ , then $\left\|\begin{array}{cc} a=2R\cos\beta\ ;\ b=2R\sin\beta \\\\ c=2R\sin\gamma\ ;\ d=2R\cos\gamma\\\\ e=2R\ ;\ f=2R\sin(\beta+\gamma )\end{array}\right\|$. In conclusion, $AB\cdot CD+AD\cdot BC=AC\cdot BD\iff$ $ac+bd=ef\iff$

$\cos\beta\sin\gamma +\sin\beta\cos\gamma=$ $\sin (\beta +\gamma )$ , what is truly . The relation $\boxed{\sin (x+y)=\sin x\cos y+\sin y\cos x}$ is named "the trigonometric form of the Ptolemy's theorem".
This post has been edited 49 times. Last edited by Virgil Nicula, Apr 12, 2017, 6:45 PM

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76. A parallelogram and a circle (nice !).
75. Extended Steinbart Theorem.
74. Inegalitatea Erdos-Mordell.
73. A nice problems with a square and a perpendicularity.
+ July 2010
72. Colinearity in a triangle - H , P , M .
70. Colinearity with pole/polar - H\in PQ .
71. An angle questions (synthetic/trigonometric proofs).
69*1. Slicing problem 3.
69. Position of x1, x2 w.r.t. a given real number k.
68. Some nice applications of "pole/polar".
67. Remarkable algebraic/geometrical inequalities (1).
66. Remarkable perpendicularities in a triangle.
65. Parallel to BC through I .
64. Problems of the type "instant" (at most one minute).
63. A nice problem with radical center (livetolove212).
62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
59. IMO 2010, problem 4.
58. A cevian and two incircles.
57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by MrMustache, Nov 11, 2020, 4:52 PM

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    by piphi, Jun 10, 2020, 11:44 PM

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  • Joined: Jun 22, 2005
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  • Total entries: 456
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