# question about triangle

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

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Can 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 $BC$? – Américo Tavares Jul 22 '11 at 10:51
yes E lie on AC and D on BC unfortunately i can't include figure because from original figure is also not included – dato datuashvili Jul 22 '11 at 10:54
I 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. – Martin Sleziak Jul 22 '11 at 11:02
yes it is correct – dato datuashvili Jul 22 '11 at 11:09
I have not understood that the bisector and median are perpendicular. – Américo Tavares Jul 22 '11 at 11:23

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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so you said that AM=MD? – dato datuashvili Jul 22 '11 at 15:14
@user3196 AM=MD holds for the same reasons as AB=BD (the trianglse AMB, AMD are congruent). – Martin Sleziak Jul 22 '11 at 15:18
i dont understand one thing why is equal ED to BE? – dato datuashvili Jul 22 '11 at 15:19
yes AM=MD i see – dato datuashvili Jul 22 '11 at 15:20
ok @Martin Sleziak thank you very much i have understood solution thanks very much – dato datuashvili Jul 22 '11 at 15:24

Hint: The triangle is not equilateral.

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and can you give me some starting point how to solve it? – dato datuashvili Jul 22 '11 at 12:24
@user3196: Strong hint given. How did the National exam go? – André Nicolas Jul 22 '11 at 13:34
in general tasks was easy but i dont know how they wrote (i had not exam ) – dato datuashvili Jul 22 '11 at 14:42

Hint: Note that $BA=BD$, so $BC=2BA$. What familiar type of triangle does this remind you of?

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my question why is that BA=BD? – dato datuashvili Jul 22 '11 at 13:43
@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.) – Martin Sleziak Jul 22 '11 at 13:48
@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. – André Nicolas Jul 22 '11 at 13:55
yes 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 <A=90 <B=60)i have supposed this then ,in ABM(m as you said is intersection point) <A=60 degree, we can find AB by the sine laws(but wait it seems that AB=BD=AD=4 but in answers no one is equal to 4 so i am confused – dato datuashvili Jul 22 '11 at 14:12
@user3196: You should also use the fact that $AD$ is median. – Martin Sleziak Jul 22 '11 at 14:19