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# Geometry problem of a triangle

Look at the figure. it is given a scalene triangle ABC. A square are drawn in each of its sides. If points P, Q and R are the centers of the square. Prove that: 1. RP perpendicular to AQ and 2. RP = AQ.

Note by Falensius Nango
4 years, 7 months ago

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## Comments

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I was expecting proofs along the lines of complex numbers of vectors, which would be considered a standard exercise. There's a more basic approach, which uses similar ideas.

Hint: Let the square be labelled $$ABDE$$. Consider triangles $$ABQ$$ and $$EBC$$. Consider triangles $$ARC$$ and $$AEC$$. Hence, we get fact 2. Show further that $$EC$$ makes a $$45^\circ$$ angle with both $$AQ$$ and $$RP$$, which gives fact 1.

Staff - 4 years, 7 months ago

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Thank u mr calvin for your hint. Just want to give a correction. Is it true that we should consider ARC and AEC, i think we should consider triangle ARP and AEC. Thank u. It was helping me a lot.

- 4 years, 7 months ago

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Here's a proof if you're familiar with complex numbers:

Let the complex numbers $$a,b,c$$ stand for the points on the triangle. Then $$\large p=c+\frac{\sqrt{2}}{2}(a-c)e^{i\frac{-\pi}{4}}=c+\frac{\sqrt{2}}{2}(a-c)(\frac{1}{\sqrt{2}}(1-i))=c+\frac{1}{2}(a-c)(1-i)$$

$$\large p=\frac{1}{2}c(1+i)+\frac{1}{2}a(1-i)$$

By analogy,

$$\large q=\frac{1}{2}b(1+i)+\frac{1}{2}c(1-i)$$

$$\large r=\frac{1}{2}a(1+i)+\frac{1}{2}b(1-i)$$

Giving,

$$\large a-q= a-\frac{1}{2}b(1+i)-\frac{1}{2}c(1-i)$$

$$\large p-r= \frac{1}{2}c(1+i)+\frac{1}{2}a(1-i)-\frac{1}{2}a(1+i)-\frac{1}{2}b(1-i)$$

$$\large = \frac{1}{2}c(1+i)-ia-\frac{1}{2}b(1-i)$$

$$\large =-ia+\frac{1}{2}b(i-1)+ \frac{1}{2}c(1+i)$$

$$\large =-ia+\frac{1}{2}b(i-1)+ \frac{1}{2}c(i+1)$$

$$\large =-i(a-\frac{1}{2}b(1+i)- \frac{1}{2}c(1-i))$$

$$\large =e^{-i\frac{\pi}{2}}(a-q)$$

QED

- 4 years, 7 months ago

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thank u...

- 4 years, 7 months ago

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