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=\(2\times OC\)
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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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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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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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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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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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Yeah, It's okay. :) – Mehul Arora · 1 year, 10 months ago
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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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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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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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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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Yeah but I was thinking about its proof!! – Ambuj Shrivastava · 1 year, 10 months ago
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You typoed the theorem. – Sharky Kesa · 1 year, 10 months ago
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Please change it, my lord. – Mehul Arora · 1 year, 10 months ago
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