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# triangle q

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$$ABCD$$ is a rectangle and $$P$$ is any point on $$AC$$ (except $$A$$ and $$C$$). Through $$P$$ is plotted a line parallel to $$BC$$ which cuts $$AB$$  and $$DC$$ at $$R$$ and $$S$$, respectively. Also, through $$S$$ is plotted a line parallel to $$AC$$ which cuts $$AD$$ at $$T$$.

Find $$\dfrac{\text{Area }TSPA}{\text{Area } PRB}$$.

Jan 15, 2021

#1
+112523
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I have not looked for a general solution but the question implies that the ratio is always the same.

If you make ABCD a square and let P be the midpoint of AC then the solution is super easy to find.

Jan 15, 2021
#2
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[TSPA] / [PRB] = 2

Jan 16, 2021
#3
+112523
+1

Refer to the diagram below:

The area of the original rectangle can be assigned any value l since we are only interested in the area ratio of the 2 internal sections.

So I have let the area be 2 units

The width is w

The length is 2/w

The original brown rectangle and the smaller red rectangle are similar figures.  So I have said the ratio of the sides is delta where delta is a positive number less than 1.

From this all the given dimensions and be found.

Are of blue parallelogram $$=\frac{2\delta}{w}*w(1-\delta)=2\delta(1-\delta)$$

Area of green triangle =$$\frac{1}{2}*\frac{2}{w}(1-\delta)*\delta w =\delta(1-\delta)$$

$$\dfrac{\text{Area }TSPA}{\text{Area } PRB} =2$$

Jan 16, 2021