suppose we have some triangle ABC with AC as base.there is BE bisector and AD median ,they intersect each other at right angle or are perpendicular,we should find lengths of triangle.we know that bisector and median are equal BE=AD=4. from my point of view at suppose level we can conclude that this triangle is equilateral ,because bisector and median are equal,they intersect at right angle or it seems they are perpendicular bisector or altitude?am i correct?also i think that key to solve such problem when there is not given additional information it to suppose such kind of possibilities,please help me to solve this problem
question about triangle
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0The triangle is not equilateral. See [this](http://calcauxprobteor.files.wordpress.com/2011/07/trianglemedianbisector.jpg?w=293&h=300) figure. – 2011-07-22
3 Answers
I think that what was provided in the discussion under @Andre's hint leads to a solution. (Probably there is a much simpler way, but I'll post my attempt anyway.)
I will denote by $M$ the intersection of $AD$ and $BC$.
Fact 1 |AB|=|BD|, |AM|=|MD| (and also |AE|=|ED|)
Fact 2 $|\triangle ABE|=|\triangle BDE|=|\triangle CDE|=\frac13|\triangle ABC|$
Fact 3 $|AE|=\frac{|AC|}3$
Fact 4 $|\triangle AME|=\frac{|\triangle AMC|}3=\frac{|\triangle ADC|}6=\frac{|\triangle ABC|}{12}=\frac{|\triangle ABE|}4$
Fact 5 $|EM|=\frac{|EB|}4=1$
As now I know all the lengths of AM, BM,DM, EM, I can use right triangles to calculate:
$|AB|=\sqrt{13}$
$|AE|=\sqrt{5}$ $\Rightarrow$ $|AC|=3\sqrt{5}$
$|BD|=\sqrt{13}$ $\Rightarrow$ $|BC|=2\sqrt{13}$
Now I can use cosine law for triangles ABE and CBE to check, whether I get the same value in both cases. (The values of the cosine should be the same, since BE is the bisector.) I get:
$\cos\frac\beta2 = \frac{13+16-5}{2.4.\sqrt{13}} = \frac{24}{8.\sqrt{13}} = \frac 3{\sqrt{13}}$
$\cos\frac\beta2 = \frac{4.13+16-4.5}{2.4.2\sqrt{13}} = \frac{48}{16.\sqrt{13}} = \frac 3{\sqrt{13}}$.
You can see why the equalities from Fact 1 hold in Andre's answer and the comments following it.
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1ok @Martin Sleziak thank you very much i have understood solution thanks very much – 2011-07-22
Hint: The triangle is not equilateral.
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0in general tasks was easy but i dont know how they wrote (i had not exam ) – 2011-07-22
Hint: Note that $BA=BD$, so $BC=2BA$. What familiar type of triangle does this remind you of?
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0sorry AE/EC=1/2 – 2011-07-22