In triangle ABC, angle C = 90 degrees, angle A = 30 degrees, and D is the foot of the altitude from C to hypotenuse AB. What is the ratio of the area of triangle BDC to the area of triangle ADC?

Guest May 24, 2020

#1**+1 **

**In triangle ABC, angle C = 90 degrees, angle A = 30 degrees, and D is the foot of the altitude from C to hypotenuse AB. **

**What is the ratio of the area of triangle BDC to the area of triangle ADC?**

\(\text{Let $\angle B = 90^\circ-30^\circ = 60^\circ$ }\)

\(\begin{array}{|rclcrcl|} \hline \tan{30^\circ} &=& \dfrac{CD}{AD} &~\qquad & \tan{60^\circ} &=& \dfrac{CD}{BD} \\ \hline \dfrac{\tan{30^\circ}}{\tan{60^\circ}} &=& \dfrac{CD}{AD}\above 1pt \dfrac{CD}{BD} \\\\ \dfrac{\tan{30^\circ}}{\tan{60^\circ}} &=& \dfrac{CD}{AD} \times \dfrac{BD}{CD} \\\\ \mathbf{\dfrac{\tan{30^\circ}}{\tan{60^\circ}}} &=& \mathbf{\dfrac{BD}{AD}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{[BDC]}{[ADC]} &=& \dfrac{BD*CD}{2}\above 1pt \dfrac{AD*CD}{2} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{BD*CD}{2} \times \dfrac{2}{AD*CD} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{BD}{AD} \quad | \quad \mathbf{\dfrac{\tan{30^\circ}}{\tan{60^\circ}}=\dfrac{BD}{AD}} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{\tan{30^\circ}}{\tan{60^\circ}} \quad | \quad \tan{30^\circ}=\dfrac{\sqrt{3}}{3},\ \tan{60^\circ}=\sqrt{3} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{\sqrt{3}}{3}\above 1pt \sqrt{3} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{\sqrt{3}}{3\sqrt{3}} \\\\ \mathbf{\dfrac{[BDC]}{[ADC]}} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

The area of triangle BDC to the area of triangle ADC is \(\mathbf{\dfrac{1}{3}}\)

heureka May 25, 2020