I made a simple configuration , whose converse is also true. Though the proof is very easy , the important thing is that its converse is also true!

**Problem statement:** Let \(AB \perp BC\) , \(DC \perp BC\). Consider a point \(P\) inside \(ABCD\). Let \(P_1\) be its reflection in \(\overline{AB}\) and \(P_2\) be its reflection in \(\overline{CD}\). Let \(\overrightarrow{P_1B} \cap \overrightarrow{P_2C} = {P_3}\). Prove that \(\overline{P_3P} \perp \overline{P_1P_2}\).

**Statement converse:** Consider a quadrilateral \(ABCD\) . Consider a point \(P\) inside \(ABCD\). Let \(P_1\) be its reflection in \(\overline{AB}\) and \(P_2\) be its reflection in \(\overline{CD}\). Let \(\overrightarrow{P_1B} \cap \overrightarrow{P_2C} ={P_3}\). If \(\overline{P_3P} \perp \overline{P_1P_2}\) , then prove that \(AB \perp BC\) , \(DC \perp BC\).

I have my own solution too. Please post awesome "complete" solutions below. Enjoy!

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## Comments

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TopNewestThe very next revolutionary book in the history of mathematics -

Nihar's treatises on Euclidean Geometry.Log in to reply

Well , I think you over-understood me as the best in geometry. I am just "good" at geometry since its one of my interests. Anyway , thanks!

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Do you have any other postulates which you have kept a secret?

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@Calvin Lin @Azhaghu Roopesh M @Trevor Arashiro @Sharky Kesa Please see my discovery. Thanks!

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Since points A and D are "essentially useless" other than saying that we have parallel / perpendicular lines, you should remove them from the statement.

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Uum ... I have points \(A,D\) just for labeling the angles , segments , that is for notation convenience. I am not able to understand why are you saying to remove them.

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Get rid of "right trapezoid such that ....", and list the important information as \( AB\perp BC \), \( BC \perp CD \).

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Congrats! xD

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It's really nice seeing you discovering postulates at such an age.

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Actually this is not a postulate. This is a simple configuration whose converse is also true. Postulates/axioms are just defined and not proved. They are "used" to prove things.

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What's a TONCAS 1?

Btw, the proof is simple for the positive statement. Just uses scale factors and similar triangles. And by proving the positive, the converse is proved here. I'm curious as to what your extension is. At any guess, I'd say reflecting a point about two parallel lines and having the three points form a right triangle about the intersection of the two lines.

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No the extension is something else. "Stay tuned" I will post it soon.

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Mm, \(k^2\). You piqued my interest.

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