# In $\Delta ABC$, angle bisector of $\angle ABC$ and median on side $BC$ intersect perpendicularly

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!

• Well aren't the questions different? Also the previous question was posted by me only. 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 $$({a,b,c}) = ({2\sqrt{13},3\sqrt{5},\sqrt{13}})$$

• Why is $BD = AB$ ? How do you show $ABO$ and $DBO$ are congruent? Oct 26 '20 at 13:17
• Angle bisector is a symmetry line in isosceles triangle. So $\triangle ABO \cong \triangle DBO$. Identifying this is quite useful in Olympiad problems. Oct 26 '20 at 13:20
• Got it, $ASA$ Congruence. Oct 26 '20 at 13:20 $$\triangle ABO \cong \triangle DBO$$ (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 $$BE = 4$$

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

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

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