How is DA equal to 1/x.

Question

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The question asked about this shape was...

"Let BP = x, and show that AB = AP = DP = 1, and DA = 1/x"

From what I extract, triangle ADB is isosceles where DA = DB, where DB = DP + BP, and from the question BP = x, and DP = 1 and therefore DA = 1 + x, not 1/x.

Is this question wrong.

Answer 1

I assumed that our pentagon is equilateral.

We have $$\Delta ADB\sim\Delta BAP,$$ which gives $$\frac{AD}{BA}=\frac{AB}{BP}$$ or $$\frac{AD}{1}=\frac{1}{x},$$ which gives what you want.

By the way, we got also $$1+x=\frac{1}{x}.$$

Answer 2

Because the parts are in a golden ratio, $\frac{1}{x} = 1+x$ in this case. That is, provided the side length of the (regular) pentagon is $1$.