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In $\Delta ABC$, $BE$ is the angle bisector of $\angle ABC$, $AD$ is the median on side $BC$. $AD$ intersects $BE$ at $O$ perpendicularly. If $AD = BE = 4$, find the lengths of each side of $\Delta ABC$.

What I Tried: At first I was having a hard time trying to make a bit of an accurate picture of the problem, and I made this :-

As solving this, I got no idea. Tried angle-chasing for example, if $\angle ABO = \angle DBO = x$ , then the green angles come to be $(90 - x)$ each, and then you have the brown angle to be $(90 + x)$ . You only get that $\Delta ABO \sim \Delta DBO$ , and that gives me no useful information for now.

I don't think I can use Pythagorean Theorem that much because except $AD = BE = 4$ , I have no other side-lengths to proceed. So right now, I am literally out of ideas.

Can anyone help me do this? Thank You!

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  • $\begingroup$ Well aren't the questions different? Also the previous question was posted by me only. $\endgroup$
    – Anonymous
    Oct 26 '20 at 12:48
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In $\triangle ABD$, $BD=AB$. $OA=OD=2$

Let $AB=c$, $AC=b$. $BC=a=2c$.

Also $OE=x$. $OB=4-x$

From $$\dfrac{AE}{CE}=\dfrac{BA}{BC}=\dfrac{1}{2}$$ $$AE = \dfrac{b}{3} , CE= \dfrac{2b}{3}$$

From Apollonius theorem,

$$ b^2 + c^2 = 2(4^2 + c^2)$$

$$ \Rightarrow b^2 - c^2 = 32$$

In right $\triangle BOD$, $$ 2^2 + (4-x)^2 = c^2$$

In right $\triangle AOE$, $$ 2^2 + x^2 = \dfrac{b^2}{9}$$

On solving, $x=1$

So $$ ({a,b,c}) = ({2\sqrt{13},3\sqrt{5},\sqrt{13}}) $$

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  • $\begingroup$ Why is $BD = AB$ ? How do you show $ABO$ and $DBO$ are congruent? $\endgroup$
    – Anonymous
    Oct 26 '20 at 13:17
  • $\begingroup$ Angle bisector is a symmetry line in isosceles triangle. So $\triangle ABO \cong \triangle DBO$. Identifying this is quite useful in Olympiad problems. $\endgroup$
    – cosmo5
    Oct 26 '20 at 13:20
  • $\begingroup$ Got it, $ASA$ Congruence. $\endgroup$
    – Anonymous
    Oct 26 '20 at 13:20
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Took your diagram and added a line for my solution.

enter image description here

$\triangle ABO \cong \triangle DBO$ (by Angle-Angle-Side)

So, $AO = OD = 2$ and $AB = BD = DC$.

Also, as $BE$ is besector of $\angle B$, $\frac{AE}{CE} = \frac{AB}{BC} = \frac{1}{2}$

Now extend line $BE$ and draw a perpendicular from point $C$ to extended line $BE$. Say it meets line $BE$ at point $F$.

Now $\triangle CEF \sim \triangle AEO$

So $\frac{EF}{CE} = \frac{OE}{AE} \implies EF = 2 OE \implies OF = 3 OE$

Also note that $\triangle BCF \sim \triangle BDO$

So $OB = OF = 3 OE; OB = 3, OE = 1 \,$ as $BE = 4$

$AB = \sqrt{OB^2 + OA^2} = \sqrt {13}$

$BC = 2 AB = 2\sqrt{13}$

$AE = \sqrt{OA^2 + OE^2} = \sqrt {5} \implies AC = 3\sqrt5$

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