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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?

May 24, 2020

#1
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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 $$\mathbf{\dfrac{1}{3}}$$ May 25, 2020