HELP Let ABC an isosceles right triangle with ABC = 90°. Consider the point P on AB such that $\frac{PA}{PB}=2$. Show that $PA\cdot BP=BQ\cdot BG$.

Question

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!

Answer 1

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 $$BM\cdot BP = BQ \cdot BG$$

Answer 2

Let AB = BC = x

M is the midpoint of BC. Hence BM = x/2

Given that PA/PB = 2 or BP/AB = 1/3 or BP = x/3

Hence BM × BP = x/2 × x/3 = x²/6 (LHS)

Now since G is the centroid, BG:GN = 2:1

Or BG:BN = 2/3 or BG = (2/3) × BN

In right angled ∆ABN, A = 45°

Therefore it is isosceles with AN = BN

Now, AN = AC/2 = √2x/2 = x/√2

Hence, BN = x/√2

BG = (2/3) × x/√2 = √2x/3

Since BN bisects B (as ∆ABN and ∆CBN are congruent by SSS case), BQ is the internal angle bisector of B in ∆BPC

So, by internal angle bisector theorem,

QC:QP = BC:BP = 3:1

So point Q divides PC in ratio 1:3

Let the origin be at B (0,0)

Then P(0,x/3) and C(x,0)

Hence Q (x/4,x/4) using section formula

BQ = √2x/4

BQ × BG = √2x/4 × √2x/3 = x²/6 (RHS)

Hence BM × BP = BQ × BG

Answer 3

As you can check in figure the statement is not correct:

$(BP=30)\times (PA=90)\neq( BG=33.94)\times( BQ=56.57)$

Answer 4

To see by elementary geometry, that $BM\cdot BP=BG\cdot BQ$, let $AM$ cross $CP$ at $D$, and join $BD$, $PG$.

Since $G$ is the centroid of $\triangle ABC$, and here$$\frac{AG}{GM}=\frac{AP}{PB}=\frac{2}{1}$$then$$PG\parallel BC$$

Since $\angle BPG$ is right, and $PG=PB=\frac{1}{2}AP$, then $B$ lies on the circle through $PGD$, and$$\triangle QBP\sim \triangle QGD$$ Further, since $\angle GDB$ is right, and$$\frac{DA}{DB}=\frac{PA}{PB}=\frac{2}{1}$$then $DP$ bisects $\angle ADB$, making $\angle GDQ=\angle MBG=45^o$, and thus$$\triangle QGD\sim\triangle MGB$$Therefore$$\triangle QBP\sim\triangle MGB$$that is$$\frac{BM}{BG}=\frac{BQ}{BP}$$and hence$$BM\cdot BP=BG\cdot BQ$$