contest-mathgeometrytriangles
Consider a $\triangle ABC$ with $BC = a$, $CA = b$, $AB = c$. Let $D$ be the midpoint of $BC$ and $E$ be the intersection of the bisector of $A$ with $BC$. The circle through $A$, $D$, $E$ meets $AC$, $AB$ again at $F$, $G$ respectively. Let $H \neq B$ be a point on $AB$ with $BG = GH$. Prove that $\triangle EBH $ and $\triangle ABC$ are similar and find the ratio of their areas.
As far as I can go, $\triangle{BAE}\sim\triangle{BDG}\sim\triangle{BCH}$, $\triangle{CEF}\sim\triangle{CAD}$…
P is not a nice point, so let's try and avoid it.
Hint: show that $JB^2=JP \times JA = JD^2$.
Hint: Show that 2 triangles are similar to prove the first.
Hint: define N in a similar manner to M
Let $DP$ intersect circle $I$ at $K$.
Notice that $RP\times PI =FP\times PE = KP\times PD$, therefore $R,K,I,D$ are co-cyclic. Also notive $IK=ID$, therefore all the red angles are equal immediately.
Now drop a perpendicular line from $A$ to BC, we have the two pink angles being equal as $ID$ is also perpendicular to $BC$.
Since the two cyan angles are equal and the ninty degree angles are equal, we have the two green angle at vertex $A$ being equal. Therefore the top red angle is equal to the top pink angle.
Look at the red and pink angles sharing edge $ID$, we know $PD$ is parallel to $AI$. Therefore $PD$ is perpendicular to $EF$.
Best Answer
$BG\cdot BA=BD\cdot BE$ $~\Longrightarrow$ $~2BG\cdot BA=2BD\cdot BE$ $~\Longrightarrow$ $~BH\cdot BA=BC\cdot BE$
thus, $\square AHEC$ is cyclic and $\triangle EBH\sim\triangle ABC$
the similarity ratio is $\dfrac{BE}{BA}=\dfrac{\left(\dfrac{ac}{b+c}\right)}{c}=\dfrac{a}{b+c}$ and then area ratio is $\dfrac{a^{2}}{(b+c)^{2}}$