Below is Al-Tusi’s proof of the planar sine law. Note, however, that the proof is made in modern mathematical notation, and thus is a sort of “translation” of Al-Tusi’s proof. Further, we take more steps in the proof for clarity’s sake, and follow the lead of [1]:

Theorem 3.1. Let ABC be a triangle with angles A, B, and C, and sides a, b, and c. Then,

sin(A) sin(B) sin (C) a b с

Proof. Let ABC be an arbitrary triangle. Prolong two arbitrary sides, say CA and BA, such that each is 60 units (60 is arbitrary, but Al-Tusi uses 60 for some reason). Let the endpoints of the prolonged side be T and D, respectively. Next, extend a line perpendicular to BC from T to a new point, K, and from A to a new point, L. Lastly, drawing a line segment AL perpendicular to BC gives us the desired setup for proving the sine law. See figure two figures below for examples using different triangles: Now, we can see that ABL and TBK are similar triangles (Similar triangles are those

with the same shape, but not necessarily the same size), and so since the ratios of the sides of similar triangles are equal, we have:

AL DF AC DC AB TB AL TK

With some manipulation, we get:

AB 60

AL TK

(AL) (60) = (AB)(TK) ⇒ (DF)(AC) = (AB)(TK) ⇒T HK Thus, finally, we have: D E DF AB AC TK LL A Figure 3: Law of sines DF TK DF DC TK DC C sin (C) b sin(B) B sinC sin B sin (C) C sin(B) b C

where c and b are the the sides opposite of the angles B and C. Hence, we are done the proof, and it is also simple to show that sin(A) is equal to the above final expression.