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!

  • $\begingroup$ Well aren't the questions different? Also the previous question was posted by me only. $\endgroup$
    – Anonymous
    Oct 26 '20 at 12:48

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

  • $\begingroup$ 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

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 $BE = 4$

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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.