In triangle $\triangle ABC$, we have $AB=8cm$, $AC=12cm$ and $BC=15cm$, $I$ the incenter. If $ID//AB$ and $IE//AC$ which is the perimeter of $IDE$?

Question

In triangle $\triangle ABC$, we have $AB=8cm$, $AC=12cm$ and $BC=15cm$, $I$ the incenter. If $ID//AB$ and $IE//AC$ which is the perimeter of $\triangle IDE$?

After drawing the shape accurately I have that the perimeter is $15$.

Since $ID//AB$, then from Thales we have that $\frac{ID}{AB}=\frac{IO}{OA}$ where $O$ is the intersection of $AI$ and $CB$. If it were the barycenter then we could use the property that $AI=2OI$, but it's the incenter. Could you please explain to me how to solve this question?

Answer 1

Hint: We have $\triangle IDE \sim \triangle ABC$ (since all of the sides are parallel), so it suffices to find the similarity ratio. It's a bit hard to do this directly by calculating the side lengths, but it's easier to think about altitudes. If $H$ is the foot from $A$ to $BC$ and $J$ is the foot from $I$ to $DE$, then $$\frac{\text{perimeter}(\triangle IDE)}{\text{perimeter}(\triangle ABC)}=\frac{IJ}{AH};$$ can you calculate $IJ$ and $AH$?

Answer 2

Notice that in $\triangle IDB$, $\angle BDI = 180^0 - \angle B$.

So $\angle IBD = \angle BID = \frac{\angle B}{2}$. So $ID = DB$. Similarly show, $IE = EC$.

So $ID + DE + EI = BC = 15$.

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If you do not notice that, here is another simple way to proceed instead of doing any calculations upfront -

Assume area of $\triangle ABC = A$. We know that $A = r \cdot \frac{P}{2} \ $ where $r$ is inradius and $P$ is the perimeter of $\triangle ABC$.

So $ \ \displaystyle r = \frac{2A}{P}$

Say altitude from vertex $A$ to $BC$ is $h$, then

$\displaystyle A = \frac{1}{2} BC \cdot h \implies h = \frac{2A}{15}$

As $\triangle ABC \sim \triangle IDE$ and also note that $r$ is the altitude from $I$ to $DE$ in $\triangle IDE$.

So, scale factor of $\triangle IDE$ to $\triangle ABC$, $k = \frac{r}{h} = \frac{15}{P}$.

Perimeter of $\triangle IDE = k \times P = 15$