# Thread: plz help me with a Similarity sum...

1. ## 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,

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

2. ## Parallelogram geometry

Hello snigdha
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,

Somebody plz help me with this sum....I had tried it a million times but to no avail....
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 $\displaystyle \vec{AB}=\vec{x}$ and $\displaystyle \vec{BC} = \vec{y}$. Then

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow$ 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?

3. ## 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 = $\displaystyle \tfrac{1}{2}$ AB and CQ = $\displaystyle \tfrac{1}{2}$ CD

$\displaystyle \therefore$ 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)

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

$\displaystyle \therefore$ AT = SC

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

$\displaystyle \therefore$ AP:PB = AT:TS

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

$\displaystyle \therefore$ AS : AC = 2 : 3