Seventy-five equivalent statements – High School Geometry

In a non-right triangle $ABC$ , let $a,b,c$ be the side-lengths, $h_a,h_b,h_c$ the altitudes, $F_a,F_b, F_c$ the feet of the altitudes from the respective vertices, $M_a,M_b,M_c$ the midpoints of sides $BC,CA,AB$ , $E_a,E_b,E_c$ the Euler points, $R$ the circumradius, $O$ the circumcenter, $N$ the nine-point center, $H$ the orthocenter, $O^c$ the reflection of $O$ over side $AB$ , $O^b$ the reflection of $O$ over side $AC$ , and $O^a$ the reflection of $O$ over side $BC$ . Then the following seventy-five statements are equivalent:

$E_c=F_c$
$AH=b$
$BH=a$
$AE_a=\frac{b}{2}$
$BE_b=\frac{a}{2}$
$E_aM_c=\frac{a}{2}$
$E_bM_c=\frac{b}{2}$
$OO^a=b$
$OO^b=a$
$CH=2h_c$
$h_a=AF_b$
$h_b=BF_a$
$AF_c=\frac{b^2}{2R}$
$BF_c=\frac{a^2}{2R}$
$\frac{a}{c} =\frac{h_c}{AF_b}$
$\frac{b}{c}=\frac{h_c}{BF_a}$
$\frac{a}{b}=\frac{BF_a}{AF_b}$
$R=\frac{b^2-a^2}{2c}$
$h_c=R\cos C$
$\cos A=\frac{b}{\sqrt{a^2+b^2}}$
$\cos B=-\frac{a}{\sqrt{a^2+b^2}}$
$\cos C=\frac{2ab}{a^2+b^2}$
$\sin A=\frac{a}{\sqrt{a^2+b^2}}$
$\sin B=\frac{b}{\sqrt{a^2+b^2}}$
$\sin C=\frac{b^2-a^2}{a^2+b^2}$
$\cos^2 A+\cos^2 B=1$
$\sin^2 A+\sin^2 B=1$
$a\cos A+b\cos B=0$
$\sin A+\cos B=0$
$\cos A-\sin B=0$
$2\cos A\cos B+\cos C=0$
$2\sin A\sin B-\cos C=0$
$\cos A\cos B+\sin A\sin B=0$
$a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C$
$\sin A\sin B=\frac{ab}{b^2-a^2}\sin C$
$\sin^2B-\sin^2A=\sin C$
$\cos^2A-\cos^2B=\sin C$
$OH^2=5R^2-c^2$
$h_a^2+h_b^2=AB^2$
$\frac{h_a}{a}+\frac{h_b}{b}=\frac{c}{h_c}$
$a^2+b^2=4R^2$
$\left(c+2AF_c\right)^2=a^2+b^2$ or $\left(c+2BF_c\right)^2=a^2+b^2$
$A-B=\pm 90^{\circ}$
$(a^2-b^2)^2=(ac)^2+(cb)^2$
$AH^2+BH^2+CH^2=8R^2-c^2$
$b=2R\cos A$
$E_a$ is the reflection of $M_b$ over side $AB$
$E_b$ is the reflection of $M_a$ over side $AB$
$\triangle ABH$ is congruent to $\triangle ABC$
$\triangle OO^aO^b$ is congruent to $\triangle ABC$
$\triangle CNO$ is isosceles with $CN=NO$
$\triangle CNH$ is isosceles with $CN=NH$
$\triangle CHO$ is right angled at $C$
$N$ is the circumcenter of $\triangle CHO$
$\triangle O^cOC$ is right-angled at $O$
$\triangle O^cHC$ is right-angled at $H$
quadrilateral $O^cOHC$ is a rectangle
the points $O^c,O,C,H$ are concyclic with $OH$ as diameter
the reflection $O^b$ of $O$ over $AC$ lies internally on $AB$
the reflection $O^a$ of $O$ over $BC$ lies externally on $AB$
radius $OC$ is parallel to side $AB$
$F_a$ is the reflection of $F_b$ over side $AB$
segment $F_aF_b$ is perpendicular to side $AB$
the nine-point center lies on $AB$
the orthic triangle is isosceles with $F_aF_c=F_bF_c$
the geometric mean theorem holds
the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
the orthocenter is a reflection of vertex $C$ over side $AB$
segment $HC$ is tangent to the circumcircle at point $C$
median $CM_c$ has the same length as the segment $HM_c$
the bisector $M_cO$ of $AB$ is tangent to the nine-point circle at $M_c$
$AF_aBF_b$ is a convex kite with diagonals $AB$ and $F_aF_b$
altitude $CF_c$ is tangent to the nine-point circle at $F_c$
chord $F_cM_c$ is a diameter of the nine-point circle
segment $HF_c$ is tangent to the nine-point circle at $F_c$ .

It will take quite some time to establish the above equivalence. In this post we select just five.

PROVE that $\cos A=\frac{b}{\sqrt{a^2+b^2}}$

implies $\sin A=\frac{a}{\sqrt{a^2+b^2}}$

This follows from the Pythagorean identity $\cos^2A+\sin^2A=1$ .

$\begin{equation*} \begin{split} \sin A&=\sqrt{1-\cos^2A}\\ &=\sqrt{1-\left(\frac{b}{\sqrt{a^2+b^2}}\right)^2}\\ &=\sqrt{\frac{a^2}{a^2+b^2}}\\ &=\frac{a}{\sqrt{a^2+b^2}} \end{split} \end{equation*}$

PROVE that $\sin A=\frac{a}{\sqrt{a^2+b^2}}$

implies $a^2+b^2=4R^2$

, where $R$

is the radius of the circumcircle of triangle $ABC$

By the extended law of sines, we have that $\sin A=\frac{a}{2R}$ . And so:

$\begin{equation*} \begin{split} \frac{a}{2R}&=\frac{a}{\sqrt{a^2+b^2}}\\ \therefore a^2+b^2&=4R^2 \end{split} \end{equation*}$

The converse also holds.

PROVE that $a^2+b^2=4R^2$

implies $\cos B=-\frac{a}{\sqrt{a^2+b^2}}$

in a non-right triangle $ABC$

We use the identity $\cos^2B+\sin^2B=1$ and the extended sine law $\sin B=\frac{b}{2R}$ :

$\begin{equation*} \begin{split} \cos B&=\pm\sqrt{1-\sin^2B}\\ &=\pm\sqrt{1-\left(\frac{b}{2R}\right)^2}\\ &=\pm\frac{a}{\sqrt{a^2+b^2}}\\ \end{split} \end{equation*}$

Since $ABC$ is non-right, take $\cos B=-\frac{a}{\sqrt{a^2+b^2}}$ .

PROVE that $\cos B=-\frac{a}{\sqrt{a^2+b^2}}$

implies $a^2+b^2=4R^2$

$\begin{equation*} \begin{split} -\sqrt{1-\sin^2B}&=-\frac{a}{\sqrt{a^2+b^2}}\\ -\sqrt{1-\left(\frac{b}{2R}\right)^2}&-\frac{a}{\sqrt{a^2+b^2}}\\ 1-\left(\frac{b}{2R}\right)^2&=\frac{a^2}{a^2+b^2}\\ \therefore a^2+b^2&=4R^2 \end{split} \end{equation*}$

PROVE that $A^2+B^2=4r^2$

implies $\sin A+\cos B=0$

From example 3 we had $\cos B=-\frac{a}{\sqrt{a^2+b^2}}$ . By the converse of example 1 $\sin A=\frac{a}{\sqrt{a^2+b^2}}$ . Thus $\sin A+\cos B=0$ .

Takeaway

In a non-right triangle $ABC$ , let $a,b,c$ be the side-lengths and $R$ the circumradius. Then the following statements are equivalent:

$a^2+b^2=4R^2$
$\cos A=\frac{b}{\sqrt{a^2+b^2}}$
$\sin A=\frac{a}{\sqrt{a^2+b^2}}$
$\cos B=-\frac{a}{\sqrt{a^2+b^2}}$
$\sin A+\cos B=0$

Task

(Late seventies) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the midpoints of sides , the Euler points, the circumradius, the circumcenter, the nine-point center, the orthocenter, the reflection of over side , the reflection of over side , and the reflection of over side . PROVE that the following seventy-eight statements are equivalent:
1. $E_c=F_c$
2. $AH=b$
3. $BH=a$
4. $AE_a=\frac{b}{2}$
5. $BE_b=\frac{a}{2}$
6. $E_aM_c=\frac{a}{2}$
7. $E_bM_c=\frac{b}{2}$
8. $E_aM_b=h_c$
9. $E_aF_b=\frac{AH}{2}$
10. $E_bF_a=\frac{BH}{2}$
11. $OO^a=b$
12. $OO^b=a$
13. $CH=2h_c$
14. $h_a=AF_b$
15. $h_b=BF_a$
16. $AF_c=\frac{b^2}{2R}$
17. $BF_c=\frac{a^2}{2R}$
18. $\frac{a}{c} =\frac{h_c}{AF_b}$
19. $\frac{b}{c}=\frac{h_c}{BF_a}$
20. $\frac{a}{b}=\frac{BF_a}{AF_b}$
21. $R=\frac{b^2-a^2}{2c}$
22. $h_c=R\cos C$
23. $\cos A=\frac{b}{\sqrt{a^2+b^2}}$
24. $\cos B=-\frac{a}{\sqrt{a^2+b^2}}$
25. $\cos C=\frac{2ab}{a^2+b^2}$
26. $\sin A=\frac{a}{\sqrt{a^2+b^2}}$
27. $\sin B=\frac{b}{\sqrt{a^2+b^2}}$
28. $\sin C=\frac{b^2-a^2}{a^2+b^2}$
29. $\cos^2 A+\cos^2 B=1$
30. $\sin^2 A+\sin^2 B=1$
31. $a\cos A+b\cos B=0$
32. $\sin A+\cos B=0$
33. $\cos A-\sin B=0$
34. $2\cos A\cos B+\cos C=0$
35. $2\sin A\sin B-\cos C=0$
36. $\cos A\cos B+\sin A\sin B=0$
37. $a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C$
38. $\sin A\sin B=\frac{ab}{b^2-a^2}\sin C$
39. $\sin^2B-\sin^2A=\sin C$
40. $\cos^2A-\cos^2B=\sin C$
41. $OH^2=5R^2-c^2$
42. $h_a^2+h_b^2=AB^2$
43. $\frac{h_a}{a}+\frac{h_b}{b}=\frac{c}{h_c}$
44. $a^2+b^2=4R^2$
45. $\left(c+2AF_c\right)^2=a^2+b^2$ or $\left(c+2BF_c\right)^2=a^2+b^2$
46. $A-B=\pm 90^{\circ}$
47. $(a^2-b^2)^2=(ac)^2+(cb)^2$
48. $AH^2+BH^2+CH^2=8R^2-c^2$
49. $b=2R\cos A$
50. $E_a$ is the reflection of $M_b$ over side $AB$
51. $E_b$ is the reflection of $M_a$ over side $AB$
52. $\triangle ABH$ is congruent to $\triangle ABC$
53. $\triangle OO^aO^b$ is congruent to $\triangle ABC$
54. $\triangle CNO$ is isosceles with $CN=NO$
55. $\triangle CNH$ is isosceles with $CN=NH$
56. $\triangle CHO$ is right angled at $C$
57. $N$ is the circumcenter of $\triangle CHO$
58. $\triangle O^cOC$ is right-angled at $O$
59. $\triangle O^cHC$ is right-angled at $H$
60. quadrilateral $O^cOHC$ is a rectangle
61. the points $O^c,O,C,H$ are concyclic with $OH$ as diameter
62. the reflection $O^b$ of $O$ over $AC$ lies internally on $AB$
63. the reflection $O^a$ of $O$ over $BC$ lies externally on $AB$
64. radius $OC$ is parallel to side $AB$
65. $F_a$ is the reflection of $F_b$ over side $AB$
66. segment $F_aF_b$ is perpendicular to side $AB$
67. the nine-point center lies on $AB$
68. the orthic triangle is isosceles with $F_aF_c=F_bF_c$
69. the geometric mean theorem holds
70. the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
71. the orthocenter is a reflection of vertex $C$ over side $AB$
72. segment $HC$ is tangent to the circumcircle at point $C$
73. median $CM_c$ has the same length as the segment $HM_c$
74. the bisector $M_cO$ of $AB$ is tangent to the nine-point circle at $M_c$
75. $AF_aBF_b$ is a convex kite with diagonals $AB$ and $F_aF_b$
76. altitude $CF_c$ is tangent to the nine-point circle at $F_c$
77. chord $F_cM_c$ is a diameter of the nine-point circle
78. segment $HF_c$ is tangent to the nine-point circle at $F_c$ .
  (Target reached! And surpassed!)
(Extra feature) If $\triangle ABC$ satisfies equation (??), PROVE that its nine-point center $N$ divides $AB$ in the ratio $|3b^2-a^2|:|b^2-3a^2|$ .