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- 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