Let $\triangle ABC$ 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 $AD/BE$ equals

We have an isosceles triangle ABC where AB = AC. When we draw altitudes AD and BE, they intersect at point O, and we're told that $\angle AOB = 105°$.

In geometry problems involving intersecting lines, we always look for vertically opposite angles first.

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Since AD and BE intersect at O, we get:

$\angle EOD = \angle AOB = 105°$ (vertically opposite angles are equal)

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Since triangle ABC is isosceles with AB = AC, the base angles are equal:

$\angle ABC = \angle ACB = x$ (let's call this angle x)

This is a fundamental property of isosceles triangles - the angles opposite the equal sides are equal.

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In quadrilateral OECD:

$\angle EOD = 105°$ (from above)

$\angle ODC = 90°$ (since AD is altitude to BC)

$\angle OCD = x$ (same as $\angle ACB$)

$\angle OEC = 90°$ (since BE is altitude to AC)

The sum of angles in any quadrilateral is 360°:

$105° + 90° + x + 90° = 360°$

$285° + x = 360°$

$x = 75°$

Therefore: $\angle ABC = \angle ACB = 75°$

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Using the angle sum property of triangles:

$\angle BAC = 180° - (75° + 75°) = 30°$

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Since $\angle AOB = 105°$ and $\angle EOD = 105°$, and these four angles around point O sum to 360°:

$\angle AOE = \angle BOD = \frac{360° - 105° - 105°}{2} = 75°$

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In triangle OBD:

$\angle EOD = 105°$, but we need $\angle BOD = 75°$

$\angle ODB = 90°$ (since AD ⊥ BC)

Therefore: $\angle OBD = 180° - 75° - 90° = 15°$

Since $\angle ABC = 75°$ and $\angle OBD = 15°$:

$\angle ABO = 75° - 15° = 60°$

In triangle BAO:

$\angle BAO = 180° - 60° - 105° = 15°$

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In triangle BAE:

$\angle BAC = 30°$ (from above)

$\angle AEB = 90°$ (since BE ⊥ AC)

$\angle ABE = 180° - 30° - 90° = 60°$

So triangle BAE is a 30°-60°-90° triangle!

This is a special right triangle with sides in the ratio 1 : √3 : 2

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In a 30°-60°-90° triangle, if the hypotenuse is h, then:

Side opposite 30° = $\frac{h}{2}$

Side opposite 60° = $\frac{h\sqrt{3}}{2}$

In triangle BAE:

AB = h (hypotenuse)

BE = $\frac{h}{2}$ (side opposite 30° angle at A)

For AD, using triangle ABD:

$\frac{AD}{AB} = \sin(75°)$

$AD = h \times \sin(75°) = h \times \cos(15°)$

Note: $\sin(75°) = \cos(15°)$ because $75° + 15° = 90°$

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$\frac{AD}{BE} = \frac{h \times \cos(15°)}{\frac{h}{2}} = 2\cos(15°)$

The key insight here was recognizing that triangle BAE is a 30°-60°-90° triangle, which gave us BE = $\frac{h}{2}$ directly, while AD required trigonometry.

The answer is $2\cos(15°)$.