Synthetic Geometry Group-Ambuj Shrivastava's Proposal

These problems are my submission to Xuming's Geometry group.These problems are taken from various sources.
- Q1 In a triangle ABC altitudes are drawn from the vertices A,B & C at points L,M & N respectively.
Prove that : ANNB\frac{AN}{NB} * BLLC\frac{BL}{LC} * CMMA\frac{CM}{MA}=1
-Q2 Prove that in any triangle 3 times the sum of squares of the length of sides of triangle is equal to 4 times the sum of squares of the length of medians of the triangle.

-Q3 In a parallelogram ABCD a line segment is drawn from D which meets AB at at F(produced).This line segment DF passes through BC at E(which is the mid point of BC) and through diagonal AC at O
Prove that:OA=2×OC2\times OC

Note by Chaitnya Shrivastava
4 years, 1 month ago

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Q1) In any triangle, a line drawn from a vertex to the opposite side is called a cevian. Thus, AL, BM and CN are cevians.

It directly follows from Ceva's theorem that ANBL×BLBL×CMMA=1\dfrac {AN}{BL} \times \dfrac {BL}{BL} \times \dfrac {CM}{MA}=1

Mehul Arora - 4 years, 1 month ago

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Its complete proof would be as follows
triangle ALB similar to triangle CNB by AA criteria . .. \Rightarrow LBNB\frac {LB}{NB} = ALCN\frac {AL}{CN} \rightarrow 1 triangle ACL similar to triangle BCM by AA criteria . \Rightarrow CMCL\frac {CM}{CL} = BMAL\frac {BM}{AL} \rightarrow 2 triangle BMA similar to triangle CNA by AA criteria .. \Rightarrow NAMA\frac {NA}{MA} = CNBM\frac {CN}{BM} \rightarrow 3
Now, On multiplying 1,2 & 3.and rearranging we get
ANNB\frac{AN}{NB} * BLLC\frac{BL}{LC} * CMMA\frac{CM}{MA}=1
As required.

Chaitnya Shrivastava - 4 years, 1 month ago

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That's how we prove Ceva's theorem. It's the same thing, although you proved Ceva's theorem. whereas I directly used it

Mehul Arora - 4 years, 1 month ago

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@Mehul Arora Sorry I didn't know about that theorem so i thought it to be a good question but still the proof might be useful.

Chaitnya Shrivastava - 4 years, 1 month ago

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@Chaitnya Shrivastava Yeah, It's okay. :)

Mehul Arora - 4 years, 1 month ago

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@Mehul Arora Am i allowed to add more questions to the set because i recently found one.

Chaitnya Shrivastava - 4 years, 1 month ago

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@Chaitnya Shrivastava Well, maybe you can add it as a different note for the benefit of everyone.

Mehul Arora - 4 years, 1 month ago

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@Mehul Arora Okay thank you.I have a doubt regarding cevea theorem can you please explain that how can we prove it in case of an ordinary cevian that is when its not an altitude, angle bisector or median.

Chaitnya Shrivastava - 4 years, 1 month ago

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@Chaitnya Shrivastava Ceva's theorem states that ANBL×BLBL×CMMA=1\dfrac {AN}{BL} \times \dfrac {BL}{BL} \times \dfrac {CM}{MA}=1 is true for any three cevians of a triangle, not only angle bisectors, or perpendiculars, or Medians

Mehul Arora - 4 years, 1 month ago

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@Mehul Arora You typoed the theorem.

Sharky Kesa - 4 years, 1 month ago

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@Sharky Kesa Please change it, my lord.

Mehul Arora - 4 years, 1 month ago

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@Mehul Arora Yeah but I was thinking about its proof!!

Chaitnya Shrivastava - 4 years, 1 month ago

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Q2) From Appolonius' theorem, We have

AB2+AC2=2(AD2+BD24)AB2+BC2=2(BE2+AD24)AC2+BC2=2(CF2+AB24)AB^2 + AC^2 = 2(AD^2+ \dfrac {BD^2}{4}) \\ AB^2 + BC^2 = 2(BE^2+ \dfrac {AD^2}{4}) \\ AC^2 + BC^2 = 2(CF^2+ \dfrac {AB^2}{4})

Adding all these, and transposing (I'm sorry for posting the incomplete solution, It's night here.)

3(AB2+BC2+CA2)=4(AD2+BE2+CF2)3 (AB^2+BC^2+CA^2)= 4(AD^2+BE^2+CF^2)

Mehul Arora - 4 years, 1 month ago

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Q3

We have ΔOAD\Delta OAD similar to ΔOCE\Delta OCE by the AAAAAA criterion. (DOA=COE\angle DOA = \angle COE, ODA=OEC\angle ODA = \angle OEC, OAD=OCE\angle OAD = \angle OCE). Furthermore, 2CE=AD2CE = AD, which implies that ΔOAD\Delta OAD is an enlargenment of ΔOCE\Delta OCE by a factor of 2. This implies that OA=2OCOA = 2OC since they are corresponding sides on the triangles. Therefore, OA=2×OCOA = 2 \times OC.

Sharky Kesa - 4 years, 1 month ago

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Nicely explained

Chaitnya Shrivastava - 4 years, 1 month ago

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