Relations between various elements of a triangle
2S = ab sin(C)
This follows from 2S = ah_{a} because
S = rp
Triangle ABC is a union of three triangles ABI, BCI, CAI, with bases
r² = p^{-1}(p - a)(p - b)(p - c)
This follows from S² = p(p - a)(p - b)(p - c) and S = rp.
1/r = 1/h_{a} + 1/h_{b} + 1/h_{c}
2S = ah_{a} = bh_{b} = ch_{c}. Therefore, a = 2S/h_{a}, etc. On the other hand, S = rp, so that
sin²(A/2) = (p - b)(p - c) / bc, etc.
First of all, sin(A) = 2·sin(A/2)cos(A/2) = 2·sin²(A/2)/tan(A/2). Therefore,
(1) | sin²(A/2) = sin(A)·tan(A/2) /2. |
We know that
(2) | sin(A) = 2S / bc |
and
(3) | tan(A/2) = r/(p - a). |
Combining (1)-(3) gives
sin²(A/2) = 2S/bc · r/(p-a) · 1/2.
Taking into account that S² = p(p - a)(p - b)(p - c) and r² = p^{-1}(p - a)(p - b)(p - c), the latter leads to
sin²(A/2) = (p - b)(p - c) / bc.
cos²(A/2) = p(p - a) / bc, etc.
Indeed, from sin²(A/2) = (p - b)(p - c) / bc,
cos²(A/2) | = 1 - sin²(A/2) | |
= 1 - (p - b)(p - c) / bc | ||
= (p(b + c) - p²) / bc | ||
= p((2p - a) - p) / bc | ||
= p(p - a) / bc. |
cos²[(C-B)/2] = [(b+c)²(p-b)(p-c)] / [a²bc]
This is the consequence of the previous two. Indeed, cos²[(C+B)/2]=sin²(A/2).
cos²[(C-B)/2]-cos²[(C+B)/2]=sin(C)sin(B)=4[ΔABC]²/(a²bc),
i.e.,
cos²[(C-B)/2]=sin²(A/2)+4p(p-a)(p-b)(p-c)/(a²bc).
In other words,
cos²[(C-B)/2]=(p-b)(p-c)/bc + 4p(p-a)(p-b)(p-c)/(a²bc)=(b+c)²(p-b)(p-c)/(a²bc).
AI² = (p - a)bc/p
Square the obvious
AI = r/sin(A/2).
Substitute there sin²(A/2) = (p - b)(p - c) / bc and
AI² | = p^{-1}(p - a)(p - b)(p - c)bc/(p - b)(p - c) |
= (p - a)bc/p. |
bc·tan(B/2)·tan(C/2)
Squaring AI = r/sin(A/2) and substituting
AI² = r²·bc/(p - b)(p - c).
By the incenter construction, tan(B/2) = r/(p - b) and also
AI² = bc·tan(B/2)·tan(C/2).
1/r = 1/r_{a} + 1/r_{b} + 1/r_{c}
S = r_{a}(p - a) = r_{b}(p - b) = r_{c}(p - c).
Therefore
1/r_{a}+ 1/r_{b} + 1/r_{c} | = (p - a)/S + (p - b)/S + (p - c)/S |
= (3p - a - b - c)/S | |
= (3p - 2p)/S | |
= p/S | |
= 1/r, |
since S = rp.
r_{a} + r_{b} + r_{c} = r + 4R
S = rp,
S = r_{a}(p - a) = r_{b}(p - b) = r_{c}(p - c).
From these we have
(4) | r_{a} + r_{b} + r_{c} - r = S(1/(p - a) + 1/(p - b) + 1/(p - c) - 1/p). |
Simple algebra yields
1/(p - a) + 1/(p - b) = c / (p - a)(p - b) and
1/(p - c) - 1/p = c / p(p - c).
And a little more effort makes a great payoff:
c / (p - a)(p - b) + c / p(p - c) = abc / p(p - a)(p - b)(p - c) = abc / S²,
by Heron's formula. To sum up, from (4)
(5) | r_{a} + r_{b} + r_{c} - r = S·abc/S² = abc / S. |
However, abc = 4RS, so that (5) implies exactly what's needed:
r_{a} + r_{b} + r_{c} - r = abc / S = 4R.
r_{a}r_{b}r_{c} = pS
S = r_{a}(p - a) = r_{b}(p - b) = r_{c}(p - c),
we immediatly obtain
r_{a}r_{b}r_{c} | = S^{3} / (p - a)(p - b)(p - c) |
= S^{3} / [S² / p], |
by Heron's formula. But
S^{3} / [S² / p] = Sp.
r+R=R(cos(A)+cos(B)+cos(C))
We know that
r + r_{c} + r_{b} - r_{a} = 4Rcos(A), r + r_{b} + r_{a} - r_{c} = 4Rcos(C), r + r_{a} + r_{c} - r_{b} = 4Rcos(B)
So that 3r+(r_{a} + r_{b} r_{c}=4R(cos(A)+cos(B)+cos(C)). But
r_{a} + r_{b} + r_{c} = r+4R
which combine into 4r+4R=4R(cos(A)+cos(B)+cos(C)), exactly as required.
r r_{a}r_{b}r_{c} = S²
This is an immediate consequence of r_{a}r_{b}r_{c} = pS and rp = S.
l_{a} = 4p(p-a)bc/(b+c)²
Follows from l_{a} = 2bc cos(A/2)/2 and cos²(A/2) = p(p - a) / bc.
l_{a} = 2bc cos(A/2)/(b+c)
Applying the sine area formula to triangles ABL_{a} and ACL_{a} and then to the entire ΔABC we see that
bl_{a}sin(A/2)/2 + cl_{a}sin(A/2)/2 = bc sin(A)/2
This simplifies to
l_{a} = bc sin(A)/ (b + c)sin(A/2) = 2bc cos(A/2) / (b + c).
m_{a}² = (b² + c²)/2 - a²/4
Let's use Stewart's theorem
AB²·DC + AC²·BD - AD²·BC = BC·DC·BD
with D being the midpoint M of BC. Then AB = c, DC = a/2, AC = b, BD = a/2, AD = m_{a}, BC = a. We have,
c²·a/2 + b²·a/2 - m_{a}²·a = a·a/2·a/2.
(The above identity could be as easily obtained with the help of the Theorem of Cosines or the Parallelogram Law; here is an example.)
abc = 4RS
Let AD be a diameter of the circumcircle of ΔABC and AH its altitude. Right triangles AHC and ABD are similar, for
AH/AC = AB/AD.
In other words,
2R·AH = AB·AC = bc.
And finally
abc = 2R·AH·a = 4RS.
bc = 2Rh_{a}
This follows from the previous derivation or by substituting
p = 4Rcos(A/2)·cos(B/2)·cos(C/2)
By the Law of Sines
a = 2R·sinA, b = 2R·sinB, c = 2R·sinC,
so that
p | = R·(sinA + sinB + sinC) |
= R·(sinA + sinB + sin(180° - A - B) | |
= R·(sinA + sinB + sin(A + B) | |
= R·(sinA + sinB + sinA·cosB + cosA·sinB) | |
= R·(sinA·(1 + cosB) + sinB·(1 + cosA)) | |
= R·(2sin(A/2)cos(A/2)·2cos²(B/2) + 2sin(B/2)cos(B/2)·2cos²(A/2)) | |
= 4R·cos(A/2)cos(B/2)(sin(A/2)cos(B/2) + sin(B/2)cos(A/2)) | |
= 4R·cos(A/2)cos(B/2)sin((A + B)/2) | |
= 4R·cos(A/2)cos(B/2)sin(90° - C/2) | |
= 4R·cos(A/2)cos(B/2)cos(C/2). |
S = 2R²sin(A)·sin(B)·sin(C)
By the Law of Sines
a = 2R·sinA, b = 2R·sinB,
For the area of the triangle we have
2S | = ab·sinC |
= 2RsinA·2RsinB·sinC | |
= 4R²·sinA·sinB·sinC. |
r = 4Rsin(A/2)·sin(B/2)·sin(C/2)
This follows directly from
cot(A/2) + cot(B/2) + cot(C/2) = cot(A/2)·cot(B/2)·cot(C/2)
This is equivalent to showing that, for A + B + C = 180°,
cos(A/2)sin(B/2)sin(C/2) + sin(A/2)cos(B/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2)
= cos(A/2)cos(B/2)cos(C/2).
Let's transform the left-hand side:
= sin((A+B)/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2).
But since (A + B)/2 C = 90° - C/2, this equals
cos(C/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2) = cos(C/2)[sin(C/2) + sin(A/2)sin(B/2)].
Reversing the steps:
sin(C/2) + sin(A/2)sin(B/2) | = cos((A+B)/2) + sin(A/2)sin(B/2) |
= cos(A/2)cos(B/2) - sin(A/2)sin(B/2) + sin(A/2)sin(B/2) | |
= cos(A/2)cos(B/2). |
Combining everything together we get the desired identity.
rR = abc / 4p
r² = p^{-1}(p - a)(p - b)(p - c) is equivalent to
r = √D / p,
where D = p(p - a)(p - b)(p - c). Also,
R = abc / 4√D.
Multiplying the two gives
rR = abc / 4p.
Note that the identity at hand also follows by combining
AH = 2R·|cos(A)|
In ΔABH, if A < 90°, ∠ABH = 90° - ∠A. (This is because ΔABH_{b} is right.) Applying the law of sines to ΔABH gives,
AH / sin(∠ABH) | = AB / sin(180° - ∠C) | |
= AB / sin(∠C) | ||
= 2R |
from the lawa of sines applied in ΔABC. Thus
2R | = AH / sin(∠ABH) | |
= AH / sin(90° - ∠A) | ||
= AH / cos(∠A), |
which proves the assertion AH = 2R·|cos(A)| when
For the case where ∠A is obtuse, H falls outside ΔABC,
p² = r_{a}r_{b} + r_{b}r_{c} + r_{c}r_{a}
As we know, r_{a}=S/(p-a). It follows that
r_{a}r_{b} + r_{b}r_{c} + r_{c}r_{a}=S²(1/[(p-b)(p-c)]+1/[(p-c)(p-a)]+1/[(p-a)(p-b)]=S²p/[(p-a)(p-b)(p-c)]=p²,
by Heron's formula.
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