2) in the opposite figure: \( \overline{A B} \) and \( \overline{A C} \) are two tangent \( m(\angle A)=50^{\circ}, m(\angle C D E)=115^{\circ} \) Prove that: 1) \( \overrightarrow{A B} \) bisect \( m(\angle A B E) \) 2) \( C B=C E \)
Solution
Step 1 ：1) Since \( \overline{A B} \) and \( \overline{A C} \) are tangents, we have \( \overline{A D} \perp \overline{A B} \) and \( \overline{A E} \perp \overline{A C} \). So, \( m(\angle ADE) = 90^{\circ} - m(\angle B) = 40^{\circ} \), and \( m(\angle AED) = 180^{\circ} - m(\angle C) = 65^{\circ} \).
Step 2 ：2) Using the angles in triangle \( \bigtriangleup ADE \), we have \( m(\angle AED) + m(\angle ADE) + m(\angle ADC) = 180^{\circ} \), so \( m(\angle ADC) = 75^{\circ} \).
Step 3 ：3) In triangle \( \bigtriangleup ADC \), we have \( m(\angle ADC) + m(\angle DCE) + m(\angle C) = 180^{\circ} \), so \( m(\angle DCE) = 35^{\circ} \), and therefore, \( m(\angle ABC) = m(\angle ABE) - m(\angle ADE) = 75^{\circ} \).
Step 4 ：4) In triangle \( \bigtriangleup ADE \), we have \( \frac{AD}{\sin(\angle ADE)} = \frac{DE}{\sin(\angle AED)} \) (by The Law of Sines), so \( \frac{CB}{\sin(40^{\circ})} = \frac{CE}{\sin(65^{\circ})} \).
Step 5 ：5) In triangle \( \bigtriangleup ACD \), we have \( \frac{CE}{\sin(40^{\circ})} = \frac{CD}{\sin(75^{\circ})} \) (by The Law of Sines), so \( \frac{CD}{\sin(75^{\circ})} = \frac{CB}{\sin(40^{\circ})} \).
Step 6 ：6) Since \( \frac{CB}{\sin(40^{\circ})} = \frac{CE}{\sin(65^{\circ})} = \frac{CD}{\sin(75^{\circ})} \), we have \( CB=CE \).
