Let $$ riangle ABC $$ 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:

Question

Let $$ 	riangle ABC $$ 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

#### Step 1: Area Relationship
The area of triangle $$ ABC $$ can be expressed in two ways using the altitudes $$ AD $$ and $$ BE $$:
$$
	ext{Area} = \frac{1}{2} 	imes AD 	imes BC = \frac{1}{2} 	imes BE 	imes AC.
$$
From this, we derive:
$$
\frac{AD}{BE} = \frac{AC}{BC}.
$$

#### Step 2: Angle Relationships
- Given $$ \angle BOA = 180^\circ - 105^\circ = 75^\circ $$, and using the sum of angles in triangle $$ AOE $$:
$$
\angle AOE + \angle AEO + \angle OAE = 180^\circ.
$$
This implies $$ \angle AEO = 15^\circ $$.

#### Step 3: Symmetry and Trigonometric Calculations
- In triangle $$ ABE $$, the altitude $$ BE = AB \cdot \cos(60^\circ) $$.
- In triangle $$ BEC $$, the altitude $$ BE = BC \cdot \cos(15^\circ) $$.

Equating the two expressions for $$ BE $$:
$$
AB \cdot \cos(60^\circ) = BC \cdot \cos(15^\circ).
$$

#### Step 4: Solve for Ratio of Altitudes
- Since $$ AB = AC $$, we substitute into the ratio:
$$
\frac{AC}{BC} = \frac{\cos(15^\circ)}{\cos(60^\circ)}.
$$
Simplify:
$$
\frac{AD}{BE} = 2\cos(15^\circ).
$$

Thus, the correct answer is **C. $$ 2\cos(15^\circ) $$**.

Answer

Answer

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