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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0Can you include a figure? I do not see how you conclude that "because bisector and median are equal,they intersect at right angle". Does $E$ lie on $AC$ and $D$ on $BC$? – 2011-07-22
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0yes E lie on AC and D on BC unfortunately i can't include figure because from original figure is also not included – 2011-07-22
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0I understood your question in such a way, that both facts about bisector and median in this triangle, i.e. they are perpendicular and they have equal length, are given in the original problem. – 2011-07-22
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0yes it is correct – 2011-07-22
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0I have not understood that the bisector and median are perpendicular. – 2011-07-22
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0here is link for problem http://www.naec.ge/images/doc/EXAMS/2011exam-matematika-13-ivlisii--.pdf see problem 34,you are right i have done badly that AC is base because triangle is given like ABC i thought that AC was base everything other is written i as wrote(it is georgian test) – 2011-07-22
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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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0so you said that AM=MD? – 2011-07-22
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1@user3196 AM=MD holds for the same reasons as AB=BD (the trianglse AMB, AMD are congruent). – 2011-07-22
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0i dont understand one thing why is equal ED to BE? – 2011-07-22
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0yes AM=MD i see – 2011-07-22
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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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0and can you give me some starting point how to solve it? – 2011-07-22
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0@user3196: Strong hint given. How did the National exam go? – 2011-07-22
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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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0my question why is that BA=BD? – 2011-07-22
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1@user3196 If $M$ denotes the intersection of $BE$ and $AD$, what can you say about the triangles $BMA$ and $BMD$? \\ Maybe I am missing something, but I do not see, how to continue after Andre's hint. (At least I do not see an easy way to continue.) – 2011-07-22
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1@user3196: Look at the two triangles that $BAD$ is divided into. They each have a right angle, and equal angles on top, so all angles correspond, the two triangles are *similar*. But the common sides are equal, so the two triangles are *congruent*. – 2011-07-22
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0yes i have seen already ,just i can't see what gives me BC=2*BA because if we say BC as leg (in right triangle) it means that – 2011-07-22
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0@user3196: You should also use the fact that $AD$ is median. – 2011-07-22
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0median we can use in fact when we are trying to define connection between areas of triangles(median divides triangle's are into two equal parts)i can't see where use median else – 2011-07-22
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0@user3196: Since AD is median, D is the midpoint of BC, which means BC=2*BD. – 2011-07-22
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0@ Martin Sleziak yes i know i said that because BA=BD it means that BA=BC/2 and i want to say this task was given to childrens in national exam(when they were passing in university)as i see something is missed in task i am thinking yet and can't find way of solution – 2011-07-22
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0also using characteristic of bisector we can say that AE/EC=1/3 – 2011-07-22
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0sorry AE/EC=1/2 – 2011-07-22