Question

# In the given figure, $$AD$$ is the bisector of the exterior $$\angle A$$ of $$\Delta ABC$$. Seg $$AD$$ intersects the side $$BC$$ produced in $$D$$. Prove that $$\displaystyle\frac{BD}{CD}=\frac{AB}{AC}$$.

Solution

## Since $$\angle CAK$$ is the exterior angle of $$\triangle ABC$$, we can say that $$m\angle CAK = \angle B + \angle C$$ (Exterior Angle Theorem).Therefore, $$m\angle CAD = \dfrac{\angle B + \angle C}2$$ (since $$AD$$ is an angle bisector)$$m\angle BAD = \angle A + \dfrac{\angle B + \angle C}2 = \dfrac{2\angle A +\angle B + \angle C}2 = \dfrac{\angle A + 180}2 = \dfrac{\angle A}2 + 90$$$$m\angle ADB = 180 - \left(\angle B + \angle BAD\right)=180-\left(\angle B + \angle A + \dfrac{\angle B + \angle C}2\right) = 180 - \angle A - \dfrac{3\angle B+\angle C}2$$Now, in $$\triangle ABD$$, applying Sine rule,$$\cfrac{BD}{\sin(\angle BAD)}= \cfrac{AB}{\sin(\angle ADB)}$$$$\cfrac{BD}{\sin\left(90+\dfrac{\angle A}2\right)} = \cfrac{AB}{\sin(\angle ADB)}$$$$\dfrac{BD}{AB} = \dfrac{\cos \left(\dfrac{\angle A}2\right)}{\sin(\angle ADB)}$$       .....$$(1)$$In $$\triangle ACD$$, applying Sine rule,$$\cfrac{CD}{\sin(\angle CAD)}=\dfrac{AC}{\sin(\angle ADB)}$$$$\cfrac{CD}{\sin\left(\dfrac{\angle B + \angle C}2\right)} = \cfrac{AC}{\sin(\angle ADB)}$$$$\dfrac{CD}{AC}=\dfrac{\sin\left(\dfrac{180-\angle A}2\right)}{\sin(\angle ADB)} = \cfrac{\cos\left(\dfrac{\angle A}2\right)}{\sin(\angle ADB)}$$    .....$$(2)$$From $$(1)$$ and $$(2)$$, we get$$\dfrac{BD}{AB} = \dfrac{CD}{AC}$$$$\Rightarrow \dfrac{BD}{CD} = \dfrac{AB}{AC}$$Hence proved.Mathematics

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