Proving Triangle Side Lengths Using Congruence and Similarity

[rushil_p]

Homework Statement

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. Homework 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.

 

[RUber]

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

 

[rushil_p]

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

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.

 

[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.
What about this:
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.

 

[SammyS]

Homework Statement

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

View attachment 87831

2. Homework 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.

You should be able to do this using similar triangles.

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

 

[rushil_p]

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.

You should be able to do this using similar triangles.

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

What about this:

As ΔAPB is congruent with ΔBQC, AP = QB
ΔANB is congruent with ΔBMC by AAS rule since ∠ANB = ∠BMC = 90°, ∠MCB = ∠NBA, AB = BC
Hence, AN = BM, the third side of the congruent triangles shown. As AN = 6 cm, BM = 6 cm
As ΔBMQ and ΔBNA are similar (scale factor 2), BN = BM*2 = 12 cm, and MN = BN - BM = 6 cm.