Question

Let $\Delta ABC$ be an isosceles triangle such that $AB$ and $AC$ are of equal length. $AD$ is the altitude from $A$ on $BC$ and $BE$ is the altitude from $B$ on $AC$. If $AD$ and $BE$ intersect at $O$ such that $\angle AOB = 105^\circ$, then $\frac{AD}{BE}$ equals

Accepted Answer

C. In $\Delta ABC$, $AB = AC$, $AD \perp BC$, and $BE \perp AC$. $AD$ bisects $\angle A$ and $BC$. In $\Delta AOB$, $\angle AOB = 105^\circ \implies \angle BOD = 180^\circ - 105^\circ = 75^\circ$. In right triangle $BOD$ (with $\angle BDO = 90^\circ$): $$\angle OBD = 90^\circ - 75^\circ = 15^\circ$$ So base angle $C = B = 90^\circ - 15^\circ = 75^\circ$. Then $\angle A = 180^\circ - 2(75^\circ) = 30^\circ$. Comparing area representations of $\Delta ABC$: $$	ext{Area} = \frac{1}{2} \cdot BC \cdot AD = \frac{1}{2} \cdot AC \cdot BE \implies \frac{AD}{BE} = \frac{AC}{BC}$$ By Sine Rule in $\Delta ABC$: $$\frac{AC}{\sin B} = \frac{BC}{\sin A} \implies \frac{AC}{BC} = \frac{\sin 75^\circ}{\sin 30^\circ} = \frac{\cos 15^\circ}{1/2} = 2\cos 15^\circ$$

Isosceles Triangle and Ratio of Altitudes

Explanation

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