To get a better idea of the community's current geometric proof "skill level," I will keep dropping the difficulty of these problems until we find an adequate starting point for the synthetic geometry group. I will appreciate any feedback on today's problem as you approach it.

Given \(\triangle ABC\), let \(D,E\) denote the midpoints of the arc \(BC\), where \(A,E\) lie on the same side \(BC\). Construct \(F\) on \(AB\) such that \(CF\perp AB\), and \(G\) on \(AE\) such that \(GF\perp DF\). Prove that \(CG=EC\).

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TopNewestHey I got around more than half way to the problem. It's almost completed. Feeling tired. I will post it tommorow. (Feeling very excited but Tired). :P :P – Surya Prakash · 2 years ago

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A slightly alternate solution:

Since \(DA\perp AG, DF\perp GF\), \(D,F,A,G\) are concyclic and thus \(\angle DGA=\angle DFB\). In addition,note that \(\angle DBA=\angle DEG\), therefore \(DEG\sim DBF\). At this point one can use the new-found information to ratio chase \(GE\) related lengths, as Surya Prakash has demonstrated, or we can continue to utilize similar (they are almost the same thing),

Since \(DEG\sim DBF\), we ask what point corresponds to \(C\) in the \(DBF\) configuration, i.e. For what point is it to \(DBF\) as \(C\) is to \(DEG\)(Not sure if this makes sense...) By some thought we quickly see that it is the midpoint of \(BC\) denoted by \(M\) (I will elaberate: it must lie on \(BC\) since \(\angle GEC=\angle FBC\), and \(M\) is the only point that satisfy \(BM=FM\)). Clearly \(DBM\sim DEC\), thus \(DEGC\sim DBFM\). SInce \(BM=FM\), we have \(CG=EC\) and are done.

Similarity is powerful isn't it? – Xuming Liang · 2 years ago

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– Surya Prakash · 2 years ago

Nice and easy one.Log in to reply

image

Hurray!!! I got the solution.

Since \(EABC\) is cyclic. It implies that \(\angle ABC = 180^{0} - \angle AEC = \angle CEG\).

We have to prove that \(CE = EG\) i.e. \(\angle CEG = \angle CGE\). It suffices to prove that \(\angle CGE = \angle ABD\).

Let '\(I\)' be foot of perpendicular from '\(C\)' onto \(EG\). But, \(\angle EFC = \angle EIC = 90^{0}\). So, \(\Delta BFC \sim \Delta CEI\). So, \(\dfrac{BF}{BC} = \dfrac{EI}{EC}\) (#)

Observe that \(\angle DBF = 180^{0} - \angle DEA = \angle DEG\) (1)

Since \(ED\) is diameter of the circle. It implies that \(\angle DAE = 90^{0}\). But \(\angle DFG = 90^{0} = \angle DAG\). So, \(A, G, D, F\) are cyclic.

It implies that \(\angle EGD = 180^{0} - \angle AFD = \angle BFD\) (2)

From (1) and (2) it implies that \(\Delta BFD \sim \Delta EGD\). So, \(\dfrac{BF}{BD} = \dfrac{EG}{ED}\)(##)

Dividing (#) by (##), we get,

\[\dfrac{BD}{BC} = \dfrac{EI}{EC} \times \dfrac{ED}{EG}\] \[\dfrac{EI}{EG} = \dfrac{BD}{BC} \times \dfrac{EC}{ED} = \dfrac{1}{2} \times \dfrac{BD}{BN} \times \dfrac{EC}{ED}\]

Since, \(\Delta ECD\) is right angled at '\(C\)' and also \(\angle CED = \angle CBD\). So, \(\cos{\angle CED} = \cos{\angle CBD}\). It implies that \(\dfrac{EC}{ED} = \dfrac{BN}{BD}\)

It gives out that \(\dfrac{EI}{EG} = \dfrac{1}{2}\) i.e. \(EG = 2 EI\). So, \(EI =GI\). It implies that \(\angle CEG = \angle CGE\) i.e. \(CG = CE\) as required.

SPECIAL THANKS TO @Xuming Liang – Surya Prakash · 2 years ago

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– Xuming Liang · 2 years ago

Congradulations :) You are the first Brilliant solver of my challenges. Great JobLog in to reply

– Nihar Mahajan · 2 years ago

Nice solution! Well Done. I have understood your solution and I realize that I was close to prove it. But I think there are few trivial typo mistakes at start that you should fix. Overall your solution is quite rich! :) :) :)Log in to reply

@Xuming Liang Do D and E lie on the circumcirlce of \(\Delta ABC\)? – Surya Prakash · 2 years ago

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– Nihar Mahajan · 2 years ago

Yes , they do.Log in to reply

So basically, we have to prove that G and E coincide, right?

@Xuming Liang – Mehul Arora · 2 years ago

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– Nihar Mahajan · 2 years ago

No. Not at all. In short you must prove that triangle CGE is isosceles.Log in to reply

Can you post the figure? – Agnishom Chattopadhyay · 2 years ago

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@Agnishom Chattopadhyay

01000011 01101111 01101110 01100111 01110010 01100001 01110100 01110011 00100000 01101111 01101110 00100000 00110001 00110000 00110000 00110000 00100000 01000110 01101111 01101100 01101111 01101111 01110111 01100101 01110010 01110011 00100001 – Mehul Arora · 2 years ago

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– Satyajit Mohanty · 2 years ago

\[\large{27741264475741985937324611021968519394650689232351903626759926561} \]Log in to reply

– Aditya Chauhan · 2 years ago

BinaryLog in to reply

– Mehul Arora · 2 years ago

Yeo. Binary it is :PLog in to reply

– Nihar Mahajan · 2 years ago

I guess this is a geometrical discussion forum... Or wait , are you thinking of computer scientific geometry? :O :3Log in to reply

– Agnishom Chattopadhyay · 2 years ago

You mean computational geometry?Log in to reply

Leave it , I was just joking :P – Nihar Mahajan · 2 years ago

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– Mehul Arora · 2 years ago

Yep. Computer scientific geom would be better :PLog in to reply

– Agnishom Chattopadhyay · 2 years ago

74 68 61 6e 6b 20 79 6f 75 21Log in to reply

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– Mehul Arora · 2 years ago

I knew you would get it in a jiffy :P :DLog in to reply

@Mehul Arora – Surya Prakash · 2 years ago

What's thatLog in to reply

– Mehul Arora · 2 years ago

I'm sure Agnishom will understand :)Log in to reply

@Mehul Arora – Surya Prakash · 2 years ago

Tell me too :PLog in to reply

– Mehul Arora · 2 years ago

Come to Slack ;)Log in to reply

– Nihar Mahajan · 2 years ago

Also , Congrats for having 1000 followers! :P :) :3Log in to reply

– Agnishom Chattopadhyay · 2 years ago

Thanks! I decided that I will be excited at first, but then again, I said, I thought I will save the excitement for 1024 followers insteadLog in to reply

– Nihar Mahajan · 2 years ago

Why 1024 only? Wait , you planning something awesome?Log in to reply

– Nihar Mahajan · 2 years ago

What difficulty are you encountering in drawing the diagram? May be I can save your time by helping you out :)Log in to reply

– Agnishom Chattopadhyay · 2 years ago

Can you help me understand how I construct the arc?Log in to reply

– Nihar Mahajan · 2 years ago

Draw circumcircle of \(\Delta ABC\). Now draw perpendicular bisector of \(BC\) and let it intersect the circumcircle at \(D,E\) and we have \(D,E\) as mid-points of arc \(BC\). Now suppose if \(A\) lies on major arc \(BC\) , then \(E\) must also lie on major arc and same if its case of minor arc.Log in to reply

@Nihar Mahajan – Surya Prakash · 2 years ago

Did u get the problemLog in to reply

– Nihar Mahajan · 2 years ago

Ah ,Don't expect me to solve Xuming's challenge so early :P I am not as awesome as he is ;)Log in to reply