A geometry problem by Yan Yau Cheng

Geometry Level 2

In △ A B C \triangle ABC , A B = 10 AB=10 and A C = 12 AC=12 , D D is a point on line B C BC such that line A D AD bisects ∠ B A C \angle BAC , The ratio B D C D \frac {BD}{CD} can be expressed as p q \frac pq , where p p and q q are positive co-prime integers, find the value of p + q p+q


The answer is 11.

Yan Yau Cheng by Yan Yau Cheng , 20, Hong Kong

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3 solutions

Krishna Ramesh
Apr 29, 2014

use the angle angle bisector theorem, which according to this question would be-

A B A C = B D C D \frac { AB }{ AC } =\frac { BD }{ CD }

⟹ \Longrightarrow 10 12 = p q \frac { 10 }{ 12 } =\frac { p }{ q } or 5 6 = p q \frac { 5 }{ 6 } =\frac { p }{ q }

so, p + q = p+q= 11 \boxed{11}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Marta Reece
Jun 12, 2017

From law of sines in triangle ACD: C D sin ⁡ α = 12 sin ⁡ β \dfrac {CD}{\sin \alpha}=\dfrac{12}{\sin\beta}

Therefore C D 12 = sin ⁡ α sin ⁡ β \dfrac{CD}{12}=\dfrac{\sin\alpha}{\sin\beta}

Likewise in triangle ABD: B D sin ⁡ α = 10 sin ⁡ ( 18 0 ∘ − β ) = 10 sin ⁡ β \dfrac {BD}{\sin \alpha}=\dfrac{10}{\sin(180^\circ-\beta)}= \dfrac{10}{\sin\beta}

Therefore B D 10 = sin ⁡ α sin ⁡ β \dfrac{BD}{10}=\dfrac{\sin\alpha}{\sin\beta}

Putting the two results together: C D 12 = B D 10 \dfrac{CD}{12}=\dfrac{BD}{10}

So that: C D B D = 12 10 = 6 5 \dfrac{CD}{BD}=\dfrac{12}{10}=\dfrac{6}{5}

And the answer is 6 + 5 = 11 6+5=\boxed{11}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Manmeswar Patnaik
Apr 29, 2014

use properties of similarities of triangles

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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