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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 : \(\frac{AN}{NB}\) * \(\frac{BL}{LC}\) * \(\frac{CM}{MA}\)=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=\(2\times OC\)

Note by Ambuj Shrivastava
1 year, 10 months ago

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Q3

We have \(\Delta OAD\) similar to \(\Delta OCE\) by the \(AAA\) criterion. (\(\angle DOA = \angle COE\), \(\angle ODA = \angle OEC\), \(\angle OAD = \angle OCE\)). Furthermore, \(2CE = AD\), which implies that \(\Delta OAD\) is an enlargenment of \(\Delta OCE\) by a factor of 2. This implies that \(OA = 2OC\) since they are corresponding sides on the triangles. Therefore, \(OA = 2 \times OC\). Sharky Kesa · 1 year, 10 months ago

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@Sharky Kesa Nicely explained Ambuj Shrivastava · 1 year, 10 months ago

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Q2) From Appolonius' theorem, We have

\(AB^2 + AC^2 = 2(AD^2+ \dfrac {BD^2}{4}) \\ AB^2 + BC^2 = 2(BE^2+ \dfrac {AD^2}{4}) \\ AC^2 + BC^2 = 2(CF^2+ \dfrac {AB^2}{4})\)

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

\(3 (AB^2+BC^2+CA^2)= 4(AD^2+BE^2+CF^2)\) Mehul Arora · 1 year, 10 months 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 \(\dfrac {AN}{BL} \times \dfrac {BL}{BL} \times \dfrac {CM}{MA}=1\) Mehul Arora · 1 year, 10 months ago

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@Mehul Arora Its complete proof would be as follows
triangle ALB similar to triangle CNB by AA criteria . .. \(\Rightarrow\) \(\frac {LB}{NB}\) = \(\frac {AL}{CN}\) \(\rightarrow\) 1 triangle ACL similar to triangle BCM by AA criteria . \(\Rightarrow\) \(\frac {CM}{CL}\) = \(\frac {BM}{AL}\) \(\rightarrow\) 2 triangle BMA similar to triangle CNA by AA criteria .. \(\Rightarrow\) \(\frac {NA}{MA}\) = \(\frac {CN}{BM}\) \(\rightarrow\) 3
Now, On multiplying 1,2 & 3.and rearranging we get
\(\frac{AN}{NB}\) * \(\frac{BL}{LC}\) * \(\frac{CM}{MA}\)=1
As required. Ambuj Shrivastava · 1 year, 10 months ago

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@Ambuj Shrivastava 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 · 1 year, 10 months 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. Ambuj Shrivastava · 1 year, 10 months ago

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@Ambuj Shrivastava Yeah, It's okay. :) Mehul Arora · 1 year, 10 months ago

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@Mehul Arora Am i allowed to add more questions to the set because i recently found one. Ambuj Shrivastava · 1 year, 10 months ago

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@Ambuj Shrivastava Well, maybe you can add it as a different note for the benefit of everyone. Mehul Arora · 1 year, 10 months 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. Ambuj Shrivastava · 1 year, 10 months ago

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@Ambuj Shrivastava Ceva's theorem states that \(\dfrac {AN}{BL} \times \dfrac {BL}{BL} \times \dfrac {CM}{MA}=1\) is true for any three cevians of a triangle, not only angle bisectors, or perpendiculars, or Medians Mehul Arora · 1 year, 10 months ago

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@Mehul Arora Yeah but I was thinking about its proof!! Ambuj Shrivastava · 1 year, 10 months ago

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@Mehul Arora You typoed the theorem. Sharky Kesa · 1 year, 10 months ago

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@Sharky Kesa Please change it, my lord. Mehul Arora · 1 year, 10 months ago

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