**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=\(2\times OC\)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestQ3

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 explainedLog 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

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

Log in to reply

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.

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply