Show that lines are parallel

**atreyyu** · April 23rd 2010, 01:58 AM

Hello

,

a random point $Q$ was chosen on side $AC$ of triangle $ABC$ . Point $P$ is the center of side $BC$ . Segments $AP$ and $BQ$ meet at $T$ . Point $R$ is the center of segment $AT$ , while point $S$ lies on $BT$ and $BS=QT$ . Show that $PS$ is parallel to $QR$ .

It's gotten too complicated for me. I'm hoping for your kind help.

**bjhopper** · April 23rd 2010, 01:39 PM

Originally Posted by atreyyu

Hello

,

a random point $Q$ was chosen on side $AC$ of triangle $ABC$ . Point $P$ is the center of side $BC$ . Segments $AP$ and $BQ$ meet at $T$ . Point $R$ is the center of $AT$ , while point $S$ belongs to $BT$ and $BS=QT$ . Show that $PS||QR$ .

It's gotten too complicated for me. I'm hoping for your kind help.

Hello atreyyu
Your problem is not clear.Suggest you post it exactly as given to you or copied from a book.

bjh

**atreyyu** · April 24th 2010, 12:43 AM

That's exactly how it was given in the book. What precisely is unclear?

**bjhopper** · April 24th 2010, 06:43 AM

Originally Posted by atreyyu

That's exactly how it was given in the book. What precisely is unclear?

Hello again atreyyu,

What does S belongs to BT mean? If i assume it means that S is the midpoint of BT I would say that PS is not parallel to QR.

bjh

**atreyyu** · April 24th 2010, 06:56 AM

Originally Posted by bjhopper

Hello again atreyyu,

What does S belongs to BT mean? If i assume it means that S is the midpoint of BT I would say that PS is not parallel to QR.

bjh

It means S is a point that was chosen on BQ such that the condition QT=BS is fulfilled.
Yeah, it's tough on me too

**running-gag** · April 24th 2010, 10:42 AM

I have checked by using coordinates and it works
There must be a geometrical solution however

**bjhopper** · April 24th 2010, 12:37 PM

Originally Posted by atreyyu

It means S is a point that was chosen on BQ such that the condition QT=BS is fulfilled.
Yeah, it's tough on me too

sorry Iwas not reading the problem correctly. Ihave tried drawing a few triangles meeting the givens and have not been able to draw one that appears to show PS parallel to QR particularly when the lines are extended. I have not been able to come up with a proof.

bjh

**atreyyu** · April 24th 2010, 12:50 PM

I did it in GeoNEXT, it's working hands down:

But then, the proof... being geometry-impaired, none of the methods I tried works...

**running-gag** · April 24th 2010, 11:37 PM

OK I have found a geometrical proof (la nuit porte conseil !

)

Here is a hint : consider point V defined as :
- V lies on (AP)
- (BV) is parallel to (AC)

Second hint : use Thalès theorem and its reciprocal

I can post my solution if you do not find

**atreyyu** · April 25th 2010, 02:12 AM

Eh, sorry. I drew it, but I still don't see it. I could use your further explanation

**Archie Meade** · April 25th 2010, 04:54 AM

Originally Posted by atreyyu

Hello

,

a random point $Q$ was chosen on side $AC$ of triangle $ABC$ . Point $P$ is the center of side $BC$ . Segments $AP$ and $BQ$ meet at $T$ . Point $R$ is the center of segment $AT$ , while point $S$ lies on $BT$ and $BS=QT$ . Show that $PS$ is parallel to $QR$ .

It's gotten too complicated for me. I'm hoping for your kind help.

Here's a possible geometric starting point,
working from the symmetry and utilising the fact that the midpoint
of two sides is chosen, causing the small triangles to be congruent.

The lines being parallel causes |BS|=|QT|.
The converse is then true.

The third side of the triangle can then be added
(left out for clarity).

**running-gag** · April 25th 2010, 04:55 AM

Here we go

Let V defined by
- V lies on (AP)
- (BV) is parallel to (AC)

To prove that (QR) and (PS) are parallel we need to show that

$\frac{TS}{TQ} = \frac{TP}{TR}$ (reciprocal of Thalès theorem)

$\frac{TS}{TQ} = \frac{TB-BS}{TQ}$

$\frac{TS}{TQ} = \frac{TB-TQ}{TQ}$ (since BS = TQ)

$\frac{TS}{TQ} = \frac{TB}{TQ} - 1$

$\frac{TS}{TQ} = \frac{TV}{TA} - 1$ (using Thalès theorem)

$\frac{TS}{TQ} = \frac{TP+PV}{TA} - 1$

$\frac{TS}{TQ} = \frac{TP+\frac{PB}{PC}PA}{TA} - 1$ (using Thalès theorem)

$\frac{TS}{TQ} = \frac{TP+PA}{TA} - 1$ (since PB = PC)

$\frac{TS}{TQ} = \frac{TP+TP+TA}{TA} - 1$

$\frac{TS}{TQ} = \frac{2TP}{TA}$

$\frac{TS}{TQ} = \frac{2TP}{2TR}$ (since R is the midpoint of [AT]

$\frac{TS}{TQ} = \frac{TP}{TR}$

**atreyyu** · April 25th 2010, 05:27 AM

Too canny. Thank you guys

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