MHB Can You Solve the Convex Quadrilateral Angle Mystery?

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In the convex quadrilateral PQRS, the sides satisfy the conditions PQ=RS and (√3+1)QR=SP, with the angle difference given by ∠RSP - ∠SPQ = 30°. The challenge is to prove that the angle difference ∠PQR - ∠QRS equals 90°. The thread encourages participation, as previous problems went unanswered. A suggested solution is available for those interested in exploring the problem further. Engaging with these geometric challenges can enhance problem-solving skills.
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Here is this week's POTW:

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In a convex quadrilateral $PQRS$, $PQ=RS$, $(\sqrt{3}+1)QR=SP$ and $\angle RSP-\angle SPQ=30^{\circ}$. Prove that $\angle PQR-\angle QRS=90^{\circ}$.

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Hello all!(Smile)

I am going to give the members another week's time to take a stab at last week's POTW. (Nod) I am looking forward with hope to receive an answer soon. (Blush)
 
No one answered last two week's POTW.(Sadface) However, for those who are interested, you can check the suggested solution by other as follows:
[TIKZ]
\coordinate[label=above: P] (P) at (2,3);
\coordinate[label=above:Q] (Q) at (4,6);
\coordinate[label=right:R] (R) at (12, 0);
\coordinate[label=below: S] (S) at (7.333333,0);
\draw (P) -- (S)-- (R)-- (Q)--(P);
\draw (P) -- (R);
\draw (Q) -- (S);
[/TIKZ]

Let $[\text{figure}]$ denote the area of figure. We have

$[PQRS]=[PQR]+[RSP]=[QRS]+[SPQ]$

Let $PQ=p,\,QR=q,\,RS=r,\,SP=s$. The above relations reduce to

$pq\sin\angle PQR+rs\sin \angle RSP=qr\sin\angle QRS+sp\sin \angle SPQ$

Using $p=r$ and $(\sqrt{3}+1)q=s$ and dividing by $pq$, we get

$\sin \angle PQR+(\sqrt{3}+1)\sin \angle RSP=\sin \angle QRS+(\sqrt{3}+1)\sin \angle SPQ$

Therefore, $\sin \angle PQR-\sin \angle QRS=(\sqrt{3}+1)(\sin \angle SPQ-\sin \angle RSP)$

This can be written in the form

$2\sin \dfrac{\angle PQR+\angle QRS}{2}\cos \dfrac{\angle PQR+\angle QRS}{2}=(\sqrt{3}+1)2\sin \dfrac{\angle SPQ-\angle RSP}{2}\cos \dfrac{\angle SPQ+\angle RSP}{2}$

Using the relations

$\cos \dfrac{\angle PQR+\angle QRS}{2}=-\cos \dfrac{\angle SPQ+\angle RSP}{2}$

and

$\cos \dfrac{\angle SPQ-\angle RSP}{2}=-\sin 15^{\circ}=- \dfrac{\sqrt{3}-1}{2\sqrt{2}}$

we obtain
$\sin \dfrac{\angle PQR-\angle QRS}{2}=(\sqrt{3}-1)\left(-\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)=\dfrac{1}{\sqrt{2}}$

This shows that

$\dfrac{\angle PQR-\angle QRS}{2}=\dfrac{\pi}{4}$ or $\dfrac{3\pi}{4}$

Using the convexity of $PQRS$, we can rule out the latter alternative. We obtain

$\angle PQR-\angle QRS=\dfrac{\pi}{2}$
 
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