Problem
GEO-B3-M06-P001 Side Through the Radius
#1
★☆☆☆☆ Level 1 of 5
In triangle \(ABC\) with circumradius \(R\), prove that \(BC=2R\sin A\).
Inspired by Prasolov trigonometric geometry method
C. Hint 1. Draw the diameter of the circumcircle through \(B\).
D. Hint 2. Use a right triangle and an inscribed angle.
Let \(BD\) be a diameter of the circumcircle. Then \(\angle BCD=90^\circ\), and \(\angle BDC=\angle BAC=A\), since both angles subtend chord \(BC\).
In right triangle \(BCD\), \(BC=BD\sin\angle BDC=2R\sin A\). This proves the claim.
A basic sine-rule problem through the circumcircle.