 #1
 18
 9
 Homework Statement:

Prove the extended Law of Sines.
Hint: Let gamma be the circumscribed circle of triangle ABC and let D be the point on gamma such that DB is a diameter of gamma. Prove that Angle BAC is congruent to Angle BDC. Use that result to prove that sin(angle BAC) = BC/2R, where R is the circumradius.
 Relevant Equations:
 Extended Law of sines: [(BC)/sin(angle BAC)] = [AC/sin(angle ABC)] = [AB/sin(angle ACB)] = 2r
I just want to know if this proof is okay, and I would like advise on how to improve it.
Proof:
i. Proving that ∠BAC ⩭∠BDC:
Let gamma be the circumscribed circle of ABC.
Let D be the point on gamma such that DB is a diameter of gamma.
The sum of the angles within a triangle equal to 180°.
Adding the angles of BAC together, b+a+c =180° where b=∠B, a=∠A, and c=∠C.
Adding the angles of BDC together, b+d+c = 180°, where b=∠B, d=∠D, and c=∠C.
Since the sum of both BAC and BDC equals 180°, b+a+c=b+d+c → a=d.
Hence, ∠BAC ⩭∠BDC.
ii. Proving that sin(∠BAC)=BC/2R:
sin(∠BAC) = P/H
= BC/BD
= BC/sin(∠BDC)
= BD
= 2R
sin(∠ADB) = P/H
= AB/BD
= AB/sin(∠ADB)
= BD
= 2R
AC/sin(∠ABC)= 2R
(BC/sin(∠BAC)) = (AC/sin(∠ABC) = (AB/sin(∠ACB)) = 2R
Proof:
i. Proving that ∠BAC ⩭∠BDC:
Let gamma be the circumscribed circle of ABC.
Let D be the point on gamma such that DB is a diameter of gamma.
The sum of the angles within a triangle equal to 180°.
Adding the angles of BAC together, b+a+c =180° where b=∠B, a=∠A, and c=∠C.
Adding the angles of BDC together, b+d+c = 180°, where b=∠B, d=∠D, and c=∠C.
Since the sum of both BAC and BDC equals 180°, b+a+c=b+d+c → a=d.
Hence, ∠BAC ⩭∠BDC.
ii. Proving that sin(∠BAC)=BC/2R:
sin(∠BAC) = P/H
= BC/BD
= BC/sin(∠BDC)
= BD
= 2R
sin(∠ADB) = P/H
= AB/BD
= AB/sin(∠ADB)
= BD
= 2R
AC/sin(∠ABC)= 2R
(BC/sin(∠BAC)) = (AC/sin(∠ABC) = (AB/sin(∠ACB)) = 2R