×

# [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, 4 months ago

Sort by:

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 · 1 year, 4 months ago

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? · 1 year, 4 months ago

Nice and easy one. · 1 year, 4 months ago

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 · 1 year, 4 months ago

Congradulations :) You are the first Brilliant solver of my challenges. Great Job · 1 year, 4 months 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! :) :) :) · 1 year, 4 months ago

@Xuming Liang Do D and E lie on the circumcirlce of $$\Delta ABC$$? · 1 year, 4 months ago

Yes , they do. · 1 year, 4 months ago

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

@Xuming Liang · 1 year, 4 months ago

No. Not at all. In short you must prove that triangle CGE is isosceles. · 1 year, 4 months ago

Can you post the figure? · 1 year, 4 months ago

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 · 1 year, 4 months ago

$\large{27741264475741985937324611021968519394650689232351903626759926561}$ · 1 year, 4 months ago

Binary · 1 year, 4 months ago

Yeo. Binary it is :P · 1 year, 4 months ago

I guess this is a geometrical discussion forum... Or wait , are you thinking of computer scientific geometry? :O :3 · 1 year, 4 months ago

You mean computational geometry? · 1 year, 4 months ago

Yeah.

Leave it , I was just joking :P · 1 year, 4 months ago

Yep. Computer scientific geom would be better :P · 1 year, 4 months ago

74 68 61 6e 6b 20 79 6f 75 21 · 1 year, 4 months ago

Comment deleted Sep 15, 2015

I knew you would get it in a jiffy :P :D · 1 year, 4 months ago

What's that @Mehul Arora · 1 year, 4 months ago

I'm sure Agnishom will understand :) · 1 year, 4 months ago

Tell me too :P @Mehul Arora · 1 year, 4 months ago

Come to Slack ;) · 1 year, 4 months ago

Also , Congrats for having 1000 followers! :P :) :3 · 1 year, 4 months 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 instead · 1 year, 4 months ago

Why 1024 only? Wait , you planning something awesome? · 1 year, 4 months ago

What difficulty are you encountering in drawing the diagram? May be I can save your time by helping you out :) · 1 year, 4 months ago

Can you help me understand how I construct the arc? · 1 year, 4 months 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. · 1 year, 4 months ago

Did u get the problem @Nihar Mahajan · 1 year, 4 months ago