# Side of a triangle (geometry)

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1. Aug 26, 2015

### rushil_p

1. The problem statement, all variables and given/known data
As shown in the diagram below, the shape consists of a square and a circle with centre Q. Given that QM = 3 cm, prove that MN = 6 cm.

Known data:
-- triangles APB and BQC are congruent
-- angle BMC = 90
-- triangles BMQ and BNA are similar and right-angled

2. Relevant equations
3. The attempt at a solution

This is a part of a structured question and I've already proved all the data listed above. I also know that AN = 6 cm as the triangles are similar, and I think that the solution might have something to do with triangle AMN being isosceles (but I can't prove that). There's probably something obvious I'm missing, so a nudge in the right direction would be greatly appreciated.

2. Aug 26, 2015

### RUber

Can you show that there is a point X such that ANMX is a square?

3. Aug 26, 2015

### rushil_p

Thank you! I think something like this would suffice: Let AX be the line parallel to MN, such that X lies on the extension of CQ. As we know ∠NMQ = 90°, ∠AXM =90° as they are interior angles between parallel lines. Since ∠MNA and ∠XAN are 90° as well, ANMX is a square, and AN = NM. Therefore MN is 6 cm.

Thanks once again.

4. Aug 26, 2015

### RUber

I think that your argument only proves that ANMX is a rectangle. I may have steered you wrong...I do not see a good way to make the conclusion along those lines.
triangles APB and BQC are congruent, so AP = AQ.
So you should be able to show that QMB = NPA.
Then use the isosceles rule to say BN = NM.
And I think you should be able to conclude the result you are looking for.

5. Aug 26, 2015

### SammyS

Staff Emeritus
You should be able to do this using similar triangles.

From the result that AN = 6cm, you can show that ΔAMN is isosceles.

6. Aug 26, 2015