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
Chaitnya Shrivastava
3 years, 3 months ago
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestQ1) 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 \(\dfrac {AN}{BL} \times \dfrac {BL}{BL} \times \dfrac {CM}{MA}=1\)
Log in to reply
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.
Log in to reply
That's how we prove Ceva's theorem. It's the same thing, although you proved Ceva's theorem. whereas I directly used it
Log in to reply
Sorry I didn't know about that theorem so i thought it to be a good question but still the proof might be useful.
Log in to reply
Yeah, It's okay. :)
Log in to reply
Am i allowed to add more questions to the set because i recently found one.
Log in to reply
Well, maybe you can add it as a different note for the benefit of everyone.
Log in to reply
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.
Log in to reply
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
Log in to reply
You typoed the theorem.
Log in to reply
Please change it, my lord.
Log in to reply
Yeah but I was thinking about its proof!!
Log in to reply
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.)
\(3 (AB^2+BC^2+CA^2)= 4(AD^2+BE^2+CF^2)\)
Log in to reply
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, \(OA = 2 \times OC\).
Log in to reply
Nicely explained
Log in to reply