# Relations between various elements of a triangle

### 2S = ab sin(C)

This follows from 2S = aha because ha = b sin(C). ### S = rp

Triangle ABC is a union of three triangles ABI, BCI, CAI, with bases AB = c, BC = a, and AC = b, respectively. The altitudes to those bases all have the length of r. ### 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/ha + 1/hb + 1/hc

2S = aha = bhb = chc. Therefore, a = 2S/ha, etc. On the other hand, S = rp, so that p = S/r, or (a + b + c) = 2S/r. From here, 2S/ha + 2S/hb + 2S/hc = 2S/r. ### 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 r² = p-1(p - a)(p - b)(p - c):

 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 sin²(A/2) = (p - b)(p - c) / bc, we obtain

AI² = r²·bc/(p - b)(p - c).

By the incenter construction, tan(B/2) = r/(p - b) and also tan(C/2) = r/(p - c). Substituting these into the above gives the required

AI² = bc·tan(B/2)·tan(C/2). ### 1/r = 1/ra + 1/rb + 1/rc

S = ra(p - a) = rb(p - b) = rc(p - c).

Therefore

 1/ra+ 1/rb + 1/rc = (p - a)/S + (p - b)/S + (p - c)/S = (3p - a - b - c)/S = (3p - 2p)/S = p/S = 1/r,

since S = rp. ### ra + rb + rc = r + 4R

S = rp,

and also

S = ra(p - a) = rb(p - b) = rc(p - c).

From these we have

 (4) ra + rb + rc - 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) ra + rb + rc - r = S·abc/S² = abc / S.

However, abc = 4RS, so that (5) implies exactly what's needed:

ra + rb + rc - r = abc / S = 4R. ### rarbrc = pS

Since

S = ra(p - a) = rb(p - b) = rc(p - c),

we immediatly obtain

 rarbrc = S3 / (p - a)(p - b)(p - c) = S3 / [S² / p],

by Heron's formula. But

S3 / [S² / p] = Sp. ### r+R=R(cos(A)+cos(B)+cos(C))

We know that

r + rc + rb - ra = 4Rcos(A), r + rb + ra - rc = 4Rcos(C), r + ra + rc - rb = 4Rcos(B)

So that 3r+(ra + rb rc=4R(cos(A)+cos(B)+cos(C)). But

ra + rb + rc = r+4R

which combine into 4r+4R=4R(cos(A)+cos(B)+cos(C)), exactly as required. ### r rarbrc = S²

This is an immediate consequence of rarbrc = pS and rp = S. ### la = 4p(p-a)bc/(b+c)²

Follows from la = 2bc cos(A/2)/2 and cos²(A/2) = p(p - a) / bc. ### la = 2bc cos(A/2)/(b+c)

Applying the sine area formula to triangles ABLa and ACLa and then to the entire ΔABC we see that

blasin(A/2)/2 + clasin(A/2)/2 = bc sin(A)/2

This simplifies to

la = bc sin(A)/ (b + c)sin(A/2) = 2bc cos(A/2) / (b + c). ### ma² = (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 = ma, BC = a. We have,

c²·a/2 + b²·a/2 - ma²·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 ∠ADB = ∠ACH. Therefore,

In other words,

2R·AH = AB·AC = bc.

And finally

abc = 2R·AH·a = 4RS. ### bc = 2Rha

This follows from the previous derivation or by substituting S = aha/2 into the final formula. ### 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:

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)
= 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 / 4D.

Multiplying the two gives

rR = abc / 4p.

Note that the identity at hand also follows by combining S = rp with abc = 4RS. ### AH = 2R·|cos(A)|

In ΔABH, if A < 90°, ∠ABH = 90° - ∠A. (This is because ΔABHb 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 A < 90°.

For the case where ∠A is obtuse, H falls outside ΔABC, ∠ABH = ∠A - 90° so at the end we'll get AH = -2R·cos(A), proving AH = 2R·|cos(A)| in this case also. ### p² = rarb + rbrc + rcra

As we know, ra=S/(p-a). It follows that

rarb + rbrc + rcra=S²(1/[(p-b)(p-c)]+1/[(p-c)(p-a)]+1/[(p-a)(p-b)]=S²p/[(p-a)(p-b)(p-c)]=p², 