Waste less time on Facebook — follow Brilliant.
×

[Solutions Posted]Geometry Proof Problem of the day #3 (Diagram Added)

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 \(CG=EC\).

Note by Xuming Liang
1 year, 10 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Hey 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 · 1 year, 10 months ago

Log in to reply

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 · 1 year, 10 months ago

Log in to reply

@Xuming Liang Nice and easy one. Surya Prakash · 1 year, 10 months ago

Log in to reply

image

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 · 1 year, 10 months ago

Log in to reply

@Surya Prakash Congradulations :) You are the first Brilliant solver of my challenges. Great Job Xuming Liang · 1 year, 10 months ago

Log in to reply

@Surya Prakash 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! :) :) :) Nihar Mahajan · 1 year, 10 months ago

Log in to reply

@Xuming Liang Do D and E lie on the circumcirlce of \(\Delta ABC\)? Surya Prakash · 1 year, 10 months ago

Log in to reply

@Surya Prakash Yes , they do. Nihar Mahajan · 1 year, 10 months ago

Log in to reply

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

@Xuming Liang Mehul Arora · 1 year, 10 months ago

Log in to reply

@Mehul Arora No. Not at all. In short you must prove that triangle CGE is isosceles. Nihar Mahajan · 1 year, 10 months ago

Log in to reply

Can you post the figure? Agnishom Chattopadhyay · 1 year, 10 months ago

Log in to reply

@Agnishom Chattopadhyay @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 · 1 year, 10 months ago

Log in to reply

@Mehul Arora \[\large{27741264475741985937324611021968519394650689232351903626759926561} \] Satyajit Mohanty · 1 year, 10 months ago

Log in to reply

@Mehul Arora Binary Aditya Chauhan · 1 year, 10 months ago

Log in to reply

@Aditya Chauhan Yeo. Binary it is :P Mehul Arora · 1 year, 10 months ago

Log in to reply

@Mehul Arora I guess this is a geometrical discussion forum... Or wait , are you thinking of computer scientific geometry? :O :3 Nihar Mahajan · 1 year, 10 months ago

Log in to reply

@Nihar Mahajan You mean computational geometry? Agnishom Chattopadhyay · 1 year, 10 months ago

Log in to reply

@Agnishom Chattopadhyay Yeah.

Leave it , I was just joking :P Nihar Mahajan · 1 year, 10 months ago

Log in to reply

@Nihar Mahajan Yep. Computer scientific geom would be better :P Mehul Arora · 1 year, 10 months ago

Log in to reply

@Mehul Arora 74 68 61 6e 6b 20 79 6f 75 21 Agnishom Chattopadhyay · 1 year, 10 months ago

Log in to reply

Comment deleted Sep 15, 2015

Log in to reply

@Agnishom Chattopadhyay I knew you would get it in a jiffy :P :D Mehul Arora · 1 year, 10 months ago

Log in to reply

@Mehul Arora What's that @Mehul Arora Surya Prakash · 1 year, 10 months ago

Log in to reply

@Surya Prakash I'm sure Agnishom will understand :) Mehul Arora · 1 year, 10 months ago

Log in to reply

@Mehul Arora Tell me too :P @Mehul Arora Surya Prakash · 1 year, 10 months ago

Log in to reply

@Surya Prakash Come to Slack ;) Mehul Arora · 1 year, 10 months ago

Log in to reply

@Agnishom Chattopadhyay Also , Congrats for having 1000 followers! :P :) :3 Nihar Mahajan · 1 year, 10 months ago

Log in to reply

@Nihar Mahajan 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 instead Agnishom Chattopadhyay · 1 year, 10 months ago

Log in to reply

@Agnishom Chattopadhyay Why 1024 only? Wait , you planning something awesome? Nihar Mahajan · 1 year, 10 months ago

Log in to reply

@Agnishom Chattopadhyay What difficulty are you encountering in drawing the diagram? May be I can save your time by helping you out :) Nihar Mahajan · 1 year, 10 months ago

Log in to reply

@Nihar Mahajan Can you help me understand how I construct the arc? Agnishom Chattopadhyay · 1 year, 10 months ago

Log in to reply

@Agnishom Chattopadhyay 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. Nihar Mahajan · 1 year, 10 months ago

Log in to reply

@Nihar Mahajan Did u get the problem @Nihar Mahajan Surya Prakash · 1 year, 10 months ago

Log in to reply

@Surya Prakash Ah ,Don't expect me to solve Xuming's challenge so early :P I am not as awesome as he is ;) Nihar Mahajan · 1 year, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...