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!

## Comments

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Nihar's treatises on Euclidean Geometry. – Swapnil Das · 1 year, 7 months agoLog in to reply

– Nihar Mahajan · 1 year, 7 months ago

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!Log in to reply

– Swapnil Das · 1 year, 7 months ago

Do you have any other postulates which you have kept a secret?Log in to reply

– Nihar Mahajan · 1 year, 7 months ago

I have one more ,I will be posting it soon on Brilliant. Actually , its the extension of this configuration.Log in to reply

– Swapnil Das · 1 year, 7 months ago

It will be a success! Bye!Log in to reply

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. – Trevor Arashiro · 1 year, 7 months ago

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– Nihar Mahajan · 1 year, 7 months ago

No the extension is something else. "Stay tuned" I will post it soon.Log in to reply

– Trevor Arashiro · 1 year, 7 months ago

Mm, \(k^2\). You piqued my interest.Log in to reply

It's really nice seeing you discovering postulates at such an age. – Swapnil Das · 1 year, 7 months ago

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– Nihar Mahajan · 1 year, 7 months ago

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.Log in to reply

Congrats! xD – Mehul Arora · 1 year, 7 months ago

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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. – Calvin Lin Staff · 1 year, 7 months ago

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– Nihar Mahajan · 1 year, 7 months ago

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.Log in to reply

– Calvin Lin Staff · 1 year, 7 months ago

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

– Nihar Mahajan · 1 year, 7 months ago

Thanks! I have edited it accordingly. Please tell more of your opinions about it (if any).Log in to reply

– Calvin Lin Staff · 1 year, 7 months ago

There is no point in saying "Join AD" is there? The idea is to remove irrelevant information like that, so that you're left with just the important attributes. This will make it easier to apply in other scenarios (where there isn't clearly a rectangular trapezoid).Log in to reply

– Nihar Mahajan · 1 year, 7 months ago

Oh , I completely understood what your intention was. I would definitely take care about this when I will post the extension of this configuration soon. "stay tuned" :PLog in to reply

– Nihar Mahajan · 1 year, 7 months ago

Oh , I see. Thanks for telling that.Lemme edit it.Log in to reply

@Calvin Lin @Azhaghu Roopesh M @Trevor Arashiro @Sharky Kesa Please see my discovery. Thanks! – Nihar Mahajan · 1 year, 7 months ago

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