Monthly Archives: March 2024

In $\triangle{ABC}$, $D$ is a point on $BC$. $\angle{ABC}=100^\circ$, $\angle{BCA}=20^\circ$, $\angle{BAD}=50^\circ$. Prove $AB=CD$. Click here for the proof. Proof 1: Clearly, $\angle{ADC}=150^\circ$, and $\angle{CAD}=10^\circ$. According to the law of sines, we have: $ \dfrac{sin\angle{BCA}}{AB}=\dfrac{sin\angle{ABC}}{AC}$ and $ \dfrac{sin\angle{CAD}}{CD}=\dfrac{sin\angle{ADC}}{AC}$ Therefore $AB=\dfrac{sin\angle{BCA}}{sin\angle{ABC}}AC=\dfrac{sin(20^\circ)}{sin(100^\circ)}AC=\dfrac{2sin(10^\circ)cos(10^\circ)}{sin(80^\circ)}AC=\dfrac{2sin(10^\circ)cos(10^\circ)}{cos(10^\circ)}AC$ $$=2sin(10^\circ)AC$$ … Continue reading →