Length of bisector (1 Viewer)

TuePhuong · Jun 9, 2016

Let ABC be a triangle having $\angle{B} = 2 \angle{C}$ , AB = 60 cm and AC = 80 cm. BD bisects $\angle{B}, D \in AC$ . Find the length of the internal angle bisector BD.

leehuan · Jun 10, 2016

TuePhuong said:
Let ABC be a triangle having

Code:

\angle{B} = 2 \angle{C}

, AB = 60 cm and AC = 80 cm. Find the length of the internal angle bisector BD.

1. You could easily use LaTeX instead of code here.
2. An assumption is made that BD bisects < ABC. What you said, whilst logically speaking implies it, does not make it clear.

$$Let $\angle ABC = 2\theta \Leftrightarrow \angle ACB = \theta$

$$In $\triangle ABC\\ \begin{align*} \frac{\sin{2\theta}}{80}&=\frac{\sin{\theta}}{60} \quad \text{(law of sines)} \\ \Rightarrow \frac{ 2\sin{\theta}\cos{\theta} }{80} &= \frac{\sin{\theta}}{60} \quad \text{ (double angle identity)} \\ \Rightarrow \cos{\theta}& =\frac{2}{3}\end{align*}$

$$In $\triangle ABD$\\ \angle ADB = 2\theta \quad \text{(exterior angle of }\triangle BCD)\\ \text{as }BD\text{ bisects }\angle ABC \Rightarrow \angle CBD = \theta \\ \Rightarrow \triangle CBD\text{ is isosceles}$

$\begin{align*}\frac{60}{\sin{2\theta}}&=\frac{AD}{\sin{\theta}} \\ \Rightarrow \frac{60}{2\sin{\theta}\cos{\theta}}&=\frac{AD}{\sin{\theta}} \\ \Rightarrow AD&=30\sec{\theta}\\& = 30 \times \frac{3}{2} \\&= 45\end{align*}$

$CD=AC-AD=80-45=35 \\ \Rightarrow BD=35 \quad \text{(matching sides on isosceles }\triangle BCD)$

I cannot guarantee the accuracy of my answer.

I also find TriABD ||| TriACB so BD/60 = CB/80. Not sure if that could produce a shortcut

TuePhuong · Jun 10, 2016

leehuan said:
1. You could easily use LaTeX instead of code here.
2. An assumption is made that BD bisects < ABC. What you said, whilst logically speaking implies it, does not make it clear.

$$Let $\angle ABC = 2\theta \Leftrightarrow \angle ACB = \theta$

$$In $\triangle ABC\\ \begin{align*} \frac{\sin{2\theta}}{80}&=\frac{\sin{\theta}}{60} \quad \text{(law of sines)} \\ \Rightarrow \frac{ 2\sin{\theta}\cos{\theta} }{80} &= \frac{\sin{\theta}}{60} \quad \text{ (double angle identity)} \\ \Rightarrow \cos{\theta}& =\frac{2}{3}\end{align*}$

$$In $\triangle ABD$\\ \angle ADB = 2\theta \quad \text{(exterior angle of }\triangle BCD)\\ \text{as }BD\text{ bisects }\angle ABC \Rightarrow \angle CBD = \theta \\ \Rightarrow \triangle CBD\text{ is isosceles}$

$\begin{align*}\frac{60}{\sin{2\theta}}&=\frac{AD}{\sin{\theta}} \\ \Rightarrow \frac{60}{2\sin{\theta}\cos{\theta}}&=\frac{AD}{\sin{\theta}} \\ \Rightarrow AD&=30\sec{\theta}\\& = 30 \times \frac{3}{2} \\&= 45\end{align*}$

$CD=AC-AD=80-45=35 \\ \Rightarrow BD=35 \quad \text{(matching sides on isosceles )}\triangle BCD$

I cannot guarantee the accuracy of my answer.

I also find TriABC ||| TriACB so BD/60 = CB/80. Not sure if that could produce a shortcut

I think you mean TriABD ||| TriACB. I have just known how to use LaTex in this forum. Thank you very much!

TuePhuong · Jun 10, 2016

leehuan said:
1. You could easily use LaTeX instead of code here.
2. An assumption is made that BD bisects < ABC. What you said, whilst logically speaking implies it, does not make it clear.

$$Let $\angle ABC = 2\theta \Leftrightarrow \angle ACB = \theta$

$$In $\triangle ABC\\ \begin{align*} \frac{\sin{2\theta}}{80}&=\frac{\sin{\theta}}{60} \quad \text{(law of sines)} \\ \Rightarrow \frac{ 2\sin{\theta}\cos{\theta} }{80} &= \frac{\sin{\theta}}{60} \quad \text{ (double angle identity)} \\ \Rightarrow \cos{\theta}& =\frac{2}{3}\end{align*}$

$$In $\triangle ABD$\\ \angle ADB = 2\theta \quad \text{(exterior angle of }\triangle BCD)\\ \text{as }BD\text{ bisects }\angle ABC \Rightarrow \angle CBD = \theta \\ \Rightarrow \triangle CBD\text{ is isosceles}$

$\begin{align*}\frac{60}{\sin{2\theta}}&=\frac{AD}{\sin{\theta}} \\ \Rightarrow \frac{60}{2\sin{\theta}\cos{\theta}}&=\frac{AD}{\sin{\theta}} \\ \Rightarrow AD&=30\sec{\theta}\\& = 30 \times \frac{3}{2} \\&= 45\end{align*}$

$CD=AC-AD=80-45=35 \\ \Rightarrow BD=35 \quad \text{(matching sides on isosceles }\triangle BCD)$

I cannot guarantee the accuracy of my answer.

I also find TriABD ||| TriACB so BD/60 = CB/80. Not sure if that could produce a shortcut

Thanks to your clue, I have another solution:
$\triangle ABD \backsim \triangle ACB \Rightarrow \frac{AB}{AC}=\frac{AD}{AB} \Rightarrow AD=\frac{AB^2}{AC}=\frac{60^2}{80}=45 \\ \Rightarrow DC=35 \Rightarrow BD = 35$ .

leehuan · Jun 10, 2016

TuePhuong said:
Thanks to your clue, I have another solution:
$\triangle ABD ||| \triangle ACB \Rightarrow \frac{AB}{AC}=\frac{AD}{AB} \Rightarrow AD=\frac{AB^2}{AC}=\frac{60^2}{80}=45 \\ \Rightarrow DC=35 \Rightarrow BD = 35$ .

That is substantially faster.

Length of bisector (1 Viewer)

TuePhuong

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leehuan

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TuePhuong

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TuePhuong

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leehuan

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