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In the figure given above, $ABPC$ is a cyclic quadrilateral and $\triangle ABC$ is an equilateral triangle. Then $\dfrac{PB}{AP} = \cdots$

A) $\dfrac{BD}{AD}\quad$ B) $\dfrac{PC}{AC}\quad$ C) $\dfrac{PD}{PC}\quad$ D) $\dfrac{PC}{AD}$

MY ATTEMPT:

I tried various methods of solving this question. At the first glance I thought that I have to apply the Ptolemy's theorem and everything should be obvious from there. So since $\triangle ABC$ is a equilateral triangle. I found out the relation that

$$PC+PB= AP$$

But after finding out this relation nothing seems very obvious what I should do next. There is no clear way there are no similar triangles I even cannot apply basic proportionality theorem (BPT).

Assistance from anyone would be greatly appreciated, whether it be in the form of hints or elaboration.

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  • $\begingroup$ Sine law may help. Note that angle PBA = $\pi/3$ + angle PBC... $\endgroup$ Commented May 29, 2024 at 13:53
  • $\begingroup$ Triangles BAP and DAB are similar $\endgroup$ Commented May 29, 2024 at 13:59
  • $\begingroup$ How can we say that ∆BAP and ∆DAB are similar and how will that help me in solving the problem ? $\endgroup$ Commented May 29, 2024 at 14:06
  • $\begingroup$ Btw $\angle BPC=120°$ if you haven't noticed $\endgroup$ Commented May 29, 2024 at 14:24
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    $\begingroup$ $APC\sim BPD$...... $\endgroup$ Commented May 29, 2024 at 15:31

1 Answer 1

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(C) $~\dfrac{PD}{PC}$ is the correct option.

In $\triangle ABP$ and $\triangle CDP$, $$\begin{aligned} &\angle APB=\angle APC=\angle DPC\qquad\text{(angles under the equal arc ${AB}$ and $AC$; equilateral $\triangle$)}\\&\angle BAP=\angle BCP=\angle DCP\qquad\text{(angles under the same arc $BP$)}\end{aligned}$$

As the sum of internal angles of a triangle is $180^\circ$, the remaining angles will be equal too. So, $\triangle ABP\sim\triangle CDP$. And we have, $$\frac{PB}{PD}=\frac{AP}{PC}\implies\color{blue}{\boxed{\frac{PB}{AP}=\frac{PD}{PC}}}$$

Also, observe that $\triangle ABP\sim\triangle ABD$. So, we have the following proportions: $$\frac{PB}{BD}=\frac{AP}{AB}\implies\color{red}{\boxed{\frac{PB}{AP}=\frac{BD}{AB}\neq\frac{BD}{AD}}}\implies\color{red}{\boxed{\frac{PB}{AP}=\frac{BD}{AC}\neq\frac{PC}{AC}}}$$

Thus, the options (A), (B), and (D) are incorrect. And the only option (C) is correct. That is, $$\color{blue}{\boxed{\frac{PB}{AP}=\frac{PD}{PC}=\frac{BD}{AB}=\frac{BD}{AC}}}$$

Hope this helps!

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  • $\begingroup$ That's a great approach of solving the problem through similarity of triangles. But I just wanted to know if this question can also be solved by the Ptolemy's theorem as my initial trials of solving this problem was based on that theorem but I failed. And one thing to note is that we never used the information in the question that ∆ABC is an equilateral triangle $\endgroup$ Commented May 30, 2024 at 13:28
  • $\begingroup$ @MINDFORGENEXUS I think Ptolemy's theorem (or its corollary) won't be helpful here. I have used the properties of the equilateral $\triangle ABC$ in the solution (like equal arcs and equal sides). $\endgroup$ Commented May 31, 2024 at 1:57
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    $\begingroup$ It turns out that there is actually a way. I consulted a math professor at my university, and after some thought, he explained an interesting relationship between the line segments PD, PB, and PC: $$1/PD = 1/PB + 1/PC$$. Remembering that I had a relation that $$PC + PB = AP$$, I realized that with a bit of algebraic manipulation of the two equations, we can show that $$PB/AP = PD/PC$$ I think you missed it . $\endgroup$ Commented Jun 2, 2024 at 14:00

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