To enjoy the geometra

Geometry Level 2

The figure below shows a right triangle A B C ABC with ∠ A = 9 0 ∘ \angle A = 90^\circ , ∠ B = 3 0 ∘ \angle B = 30^\circ and ∠ C = 6 0 ∘ \angle C = 60^\circ . A D AD is a bisector of ∠ A \angle A , B E BE is a bisector of ∠ B \angle B . If B E A D = n \dfrac {BE}{AD} = n , where n n is a positive integer, find n n ?


The answer is 2.

Aziz Alasha by Aziz Alasha , 59, Sudan

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1 solution

Chew-Seong Cheong
Jul 31, 2017

We note that ∠ A D B = ∠ C A D + ∠ A C B = 4 5 ∘ + 6 0 ∘ = 10 5 ∘ \angle ADB = \angle CAD + \angle ACB = 45^\circ + 60^\circ = 105^\circ . Using sine rule , A D A B = sin ⁡ 3 0 ∘ sin ⁡ 10 5 ∘ \dfrac {AD}{AB} = \dfrac {\sin 30^\circ}{\sin 105^\circ} ⟹ A D = sin ⁡ 3 0 ∘ sin ⁡ 10 5 ∘ A B \implies AD = \dfrac {\sin 30^\circ}{\sin 105^\circ} AB . We note that A B B E = cos ⁡ 1 5 ∘ \dfrac {AB}{BE} = \cos 15^\circ ⟹ B E = A B cos ⁡ 1 5 ∘ \implies BE = \dfrac {AB}{\cos 15^\circ} . Therefore,

B E A D = sin ⁡ 10 5 ∘ sin ⁡ 3 0 ∘ cos ⁡ 1 5 ∘ = cos ⁡ ( 9 0 ∘ − 10 5 ∘ ) sin ⁡ 3 0 ∘ cos ⁡ 1 5 ∘ = cos ⁡ ( − 1 5 ∘ ) sin ⁡ 3 0 ∘ cos ⁡ 1 5 ∘ = cos ⁡ 1 5 ∘ sin ⁡ 3 0 ∘ cos ⁡ 1 5 ∘ = 1 sin ⁡ 3 0 ∘ = 2 \begin{aligned} \frac {BE}{AD} & = \frac {\sin 105^\circ}{\sin 30^\circ \cos 15^\circ} \\ & = \frac {\cos (90^\circ - 105^\circ)}{\sin 30^\circ \cos 15^\circ} \\ & = \frac {\cos (- 15^\circ)}{\sin 30^\circ \cos 15^\circ} \\ & = \frac {\cos 15^\circ}{\sin 30^\circ \cos 15^\circ} \\ & = \frac 1{\sin 30^\circ} \\ & = \boxed{2} \end{aligned}

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