Let ABC be an isosceles right triangle with ABC = 90°, M and N the midpoints of sides BC and AC, and G is BN intersected by AM. Consider the point P on AB such that $\frac{BM}{BP}=2$. If BN intersects CP in Q, show that $PA\cdot BP=BQ\cdot BG$.
My ideas:
My drawing
Okay, so I thought of many things:
1)
MN being middle line
Maybe we can use the middle line propriety somewhere.
2)
Center of gravity G
If we take a closer look at the drawing we observe that G is the crossing of two medians.
From this, we can write that :
$BG=\frac{2\cdot BN}{3}$
$PB=\frac{AB}{3}$
ALSO
$BM=\frac{BN}{2}$
From what I wrote above we can write that:
$PB\cdot BM= \frac{AB}{3}\cdot\frac{AB}{2}$
I think we can notate AB as x and then show that $BQ\cdot BG$ equals it.
3)
The Thales Theorem
We can apply The Thales Theorem in the ABM triangle => PG is parallel with BM.
I don't know what to do forward. Hope one of you can help me. Any ideas are welcome! Thank you!
Best Answer
The title says $PA\cdot BP=BQ\cdot BG$, but I don't think this holds.
Freddy proved that $BM\cdot BP=BQ\cdot BG$ which can be proved without using coordinate geometry or vectors.
Using Menelaus's theorem (line $CQP$ passes inside $\triangle{ABN}$), we have $$\frac{AP}{PB}\times\frac{BQ}{QN}\times\frac{NC}{CA}=1$$ which implies $BQ=QN$, so $Q$ is the midpoint of the line segment $BN$.
Let $BM=a$. Then, since $BP=\frac{2a}{3},BQ=\frac{\sqrt 2}{2}a$ and $BG=\frac{2\sqrt 2}{3}a$, we have $$BM\cdot BP = BQ \cdot BG$$