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**Bhavik jain****Guest**

If P is the midpoint of the side BC of ▲ABC and Q is the midpoint of AP and if BQ produced meets AC in R prove that BQ = 3QR

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,368

hi Bhavik jain

Welcome to the forum.

No 'Euclidean' way has jumped into my head yet, but I can show you a way using vector geometry.

Post back if you are interested. Say if you have seen vector geometry used to prove triangle properties. It will tell me how much detail to put in.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,368

Will do, but I need to know how far back to go.

Do you know what a vector is?

Have you used a vector approach to find triangle properties? eg. That the 3 medians go though a single point.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,368

Ok Srikantan,

Here we go.

I'll use bold letters for the vectors.

First choose an origin. Any point will work but as the question is about the line BQR I'll choose B.

And for a 2-D problem you need two base vectors. Any will do providing they are not parallel so I choose **BA** = **a** and **BC** = **c**

The position of every other point can be written as some combination of an amount in the **a** direction and an amount in the **c** direction.

P is the midpoint of BC so **BP** = 0.5**c**

And **AP** = **AB** + **BP** = -**a** + 0.5**c**

=> **AQ** = 0.5(**AP**) = -0.5**a** + 0.25**c**

=> **BQ** = **BA** + **AQ** = **a** -0.5**a** + 0.25**c** = 0.5**a** +0.25**c**

**AC** = **AB** + **BC** = -**a** + **c**

Now we know R lies on AC but we don't know where, so suppose it is fraction f, along from A (normally I'd use a Greek letter lambda here but it gets too complicated to try and show that)

=> **AR** = f( -**a** + **c**) = -f**a** + f**c**

=> **BR** = **BA** +**AR** = **a** -f**a** + f**c** = (1-f)**a** + f**c** ......(1)

WE also know that R is some way along BQ, let's say g along where g is a number over 1.

=> **BR** = g (0.5**a** +0.25**c**) = 0.5g**a** + 0.25g**c** .......(2)

We now have two ways to describe **BR** so they must be equal. In vectors, each component must be equal so we have a pair of simultaneous equations:

**a** components: 1-f = 0.5g

**c** components: f = 0.25g

adding gives 1 = 0.75g so g = 4/3 and so f = 1/3

In (1) gives **BR** = 2/3. **a** + 1/3. **c** and in (2) gives **BR** = 2/3. **a** + 1/3. **c**

I've worked out both ways to check for mistakes. I did find one, so these are the corrected values.

As g = 4/3, we know BQ:BR = 1:4/3 = 3:4 => 3.QR = BQ as required.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,368

I have a 'Euclidean' method.

I have extended BA to D, BQR to S and constructed CSD parallel to PA.

As BP = PC the triangle BCD is similar and twice the size of BPA.

Therefore if AQ = QP = x, then SC = 2x

In triangles AQR and CSR, AP is parallel to CS, AR and RC are part of the same line, and similarly QR and RS, so these triangles are similar. As CS = twice AQ then RS is twice RQ.

Let BQ = 3y = QS. R divides QS in the ratio 1:2 then QR = y and RS = 2y.

thus BQ = 3.QR

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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