**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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## Comments

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TopNewestQ1) 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 \(\dfrac {AN}{BL} \times \dfrac {BL}{BL} \times \dfrac {CM}{MA}=1\)

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Its complete proof would be as followstriangle ALB similar to triangle CNB by AA criteria. .. \(\Rightarrow\) \(\frac {LB}{NB}\) = \(\frac {AL}{CN}\) \(\rightarrow\) 1triangle ACL similar to triangle BCM by AA criteria. \(\Rightarrow\) \(\frac {CM}{CL}\) = \(\frac {BM}{AL}\) \(\rightarrow\) 2triangle BMA similar to triangle CNA by AA criteria.. \(\Rightarrow\) \(\frac {NA}{MA}\) = \(\frac {CN}{BM}\) \(\rightarrow\) 3Now, On multiplying 1,2 & 3.and rearranging we get

\(\frac{AN}{NB}\) * \(\frac{BL}{LC}\) * \(\frac{CM}{MA}\)=1

As required.

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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

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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.)

\(3 (AB^2+BC^2+CA^2)= 4(AD^2+BE^2+CF^2)\)

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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, \(OA = 2 \times OC\).

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Nicely explainedLog in to reply