Geometry
Triangle $ABC$ has $AB = 4\sqrt{5}, BC = 2\sqrt{5}, AC = 10$. Internal bisector of $\angle ABC$ intersects $AC$ at $D$. $AC$ is extended up to $E$ such that $BD=CE$. The area of Triangle $BCE$ can be expressed as $\dfrac{a\sqrt{b}}{c}$, where $b$ and $c$ are co primes. Find $a + b + c.$ :D
Last edited by tanmoy on Tue Aug 17, 2021 11:46 pm, edited 1 time in total.
Reason: Latexed
Reason: Latexed
- gwimmy(abid)
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Re: Geometry
$AB$, $AC$ and $BC$ are the vertices of a right angled triangle. Since $BD$ is the bisector of $\angle ABC$, $BC:AB = CD:AD = 1:2$. Thus $CD = 4$ and $AD = 6$. Using the Stewart's theorem, we have, $10(BD^2+4\cdot 6) = 2 \sqrt {5} \cdot 6 + 4 \sqrt{5}\cdot 4$. Solving this equation, we get $CE = BD = 2 \sqrt {5}$. Now we extend $AB$ to $E_1$ such that, $\angle BE_1E = 90^{\circ}$. $\triangle AE_1E$ is similar to $\triangle ABC$ with $AB:EE_1 = AC : CE = 10 : 2\sqrt {5}$. Thus height of $\triangle BCE$ is $EE_1=4$ and base is $BC = 2\sqrt{5}$. Thus area of the triangle is $4\sqrt{5}$.
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Umm....the healer needs healing...
Re: Geometry
You are wrong.gwimmy(abid) wrote: ↑Thu Apr 08, 2021 1:07 pm$AB$, $AC$ and $BC$ are the vertices of a right angled triangle. Since $BD$ is the bisector of $\angle ABC$, $BC:AB = CD:AD = 1:2$. Thus $CD = 4$ and $AD = 6$. Using the Stewart's theorem, we have, $10(BD^2+4\cdot 6) = 2 \sqrt {5} \cdot 6 + 4 \sqrt{5}\cdot 4$. Solving this equation, we get $CE = BD = 2 \sqrt {5}$. Now we extend $AB$ to $E_1$ such that, $\angle BE_1E = 90^{\circ}$. $\triangle AE_1E$ is similar to $\triangle ABC$ with $AB:EE_1 = AC : CE = 10 : 2\sqrt {5}$. Thus height of $\triangle BCE$ is $EE_1=4$ and base is $BC = 2\sqrt{5}$. Thus area of the triangle is $4\sqrt{5}$.
In the second line, you said $BC : AB = CD : AD = 1 : 2$. It's right. But, in the next line, you made the mistake. Though it is a very small mistake, it made your answer wrong. Hope, you will get it.
Thank you
Last edited by tanmoy on Tue Aug 17, 2021 11:47 pm, edited 1 time in total.
Reason: Latexed
Reason: Latexed