# plz help me with a Similarity sum...

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• Mar 3rd 2009, 02:46 AM
snigdha
plz help me with a Similarity sum...
In a parallelogram ABCD, P & Q are mid-points of AB and DC respectively.
AB//PQ//DC. Join diagonal AC. AC and PQ intersect at R. Join BQ. RC and BQ intersect at S. If AS= 20cm, find,
1. triangle CQR: triangle SCQ,
2. triangle SCQ:triangle RSQ,
3. quadrilateral ASQD: //gm ABCD.

Somebody plz help me with this sum...I had tried it a million times but to no avail....
• Mar 8th 2009, 10:44 AM
Grandad
Parallelogram geometry
Hello snigdha
Quote:

Originally Posted by snigdha
In a parallelogram ABCD, P & Q are mid-points of AB and DC respectively.
AB//PQ//DC. Join diagonal AC. AC and PQ intersect at R. Join BQ. RC and BQ intersect at S. If AS= 20cm, find,
1. triangle CQR: triangle SCQ,
2. triangle SCQ:triangle RSQ,
3. quadrilateral ASQD:parallelogram ABCD.

Somebody plz help me with this sum.(Surprised)...I had tried it a million times but to no avail....(Doh)

I assume that the bit I've marked in red should be AD//PQ//BC.

Do you know how to use vectors to solve problems like this? If you do, it's quite easy. Let $\vec{AB}=\vec{x}$ and $\vec{BC} = \vec{y}$. Then

$\vec{AC} = \vec{AB} + \vec{BC} = \vec{x} + \vec{y}$

$\Rightarrow \vec{AS} = p(\vec{x} + \vec{y})$, for some number $p$

And $\vec{BQ} = \vec{BC} + \vec{CQ} = \vec{y} - \tfrac{1}{2}\vec{x}$, since Q is the mid-point of DC.

$\Rightarrow \vec{BS} = q(\vec{y} - \tfrac{1}{2}\vec{x})$, for some number $q$

But $\vec{BS} = \vec{BA} + \vec{AS} = -\vec{x} +p(\vec{x} + \vec{y}) = (p-1)\vec{x} + p\vec{y}$

$\Rightarrow (p-1)\vec{x} + p\vec{y} = q(\vec{y} - \tfrac{1}{2}\vec{x})$

Compare coefficients: $\vec{y}: p = q; \vec{x}: p-1 = -\tfrac{1}{2}q$

$\Rightarrow p = q = \tfrac{2}{3}$

So AS = $\tfrac{2}{3}$AC.

Also, R is the mid-point of AC (since triangles APR and CQR are congruent), and so RC = $\tfrac{1}{2}$AC $\Rightarrow$ RC : SC = 3 : 2

$\Rightarrow$ area triangle CQR : area triangle SCQ = 3:2, since they share the same height.

Do you want to see if you can finish it now?

Grandad
• Mar 9th 2009, 07:23 AM
Grandad
Geometry Proof
Hello snigdha

Here's a more traditional proof that the ratio AS:SC = 2:3.

See the attached diagram.

PD meets AC at T. We prove first that triangles APD, CQB are congruent.

Proof:

AD = BC (opposite sides of a parallelogram)

Angle A = angle C (opposite angles of a parallelogram)

AB = DC (opposite sides of a parallelogram)

But AP = $\tfrac{1}{2}$ AB and CQ = $\tfrac{1}{2}$ CD

$\therefore$ AP = CQ

So the triangles APD, CQB are congruent (2 sides and included angle)

Now we prove that triangles APT, CQS are congruent.

Proof:

angle APT = angle CQS (angles in congruent triangle)

and angle PAT = angle QCS (alternate angles, AB //DC)

and AP = QC (proven)

$\therefore$ triangles APT, CQS are congruent (2 angles and corresponding side)

$\therefore$ AT = SC

Then consider the triangles APT and ABS. These are similar (corresponding angles equal).

$\therefore$ AP:PB = AT:TS

But AP = PB $\therefore$ AT = TS = SC

$\therefore$ AS : AC = 2 : 3

Grandad