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= $2\times OC$

#Synthetic

$</code> ... <code>$ $< / c o d e > . . . < c o d e >$ ."> Easy Math Editor

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Sort by:

Top Newest

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 - 3 years, 10 months ago

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.

Chaitnya Shrivastava - 3 years, 10 months ago

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 - 3 years, 10 months ago

@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.

Chaitnya Shrivastava - 3 years, 10 months ago

@Chaitnya Shrivastava – Yeah, It's okay. :)

Mehul Arora - 3 years, 10 months ago

@Mehul Arora – Am i allowed to add more questions to the set because i recently found one.

Chaitnya Shrivastava - 3 years, 10 months ago

@Chaitnya Shrivastava – Well, maybe you can add it as a different note for the benefit of everyone.

Mehul Arora - 3 years, 10 months ago

@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.

Chaitnya Shrivastava - 3 years, 10 months ago

@Chaitnya 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 - 3 years, 10 months ago

@Mehul Arora – You typoed the theorem.

Sharky Kesa - 3 years, 10 months ago

@Sharky Kesa – Please change it, my lord.

Mehul Arora - 3 years, 10 months ago

@Mehul Arora – Yeah but I was thinking about its proof!!

Chaitnya Shrivastava - 3 years, 10 months ago

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 - 3 years, 10 months ago

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 - 3 years, 10 months ago

Nicely explained

Chaitnya Shrivastava - 3 years, 10 months ago

Excel in math and science

Master concepts by solving fun, challenging problems.

It's hard to learn from lectures and videos

Learn more effectively through short, interactive explorations.

Used and loved by over 7 million people

Learn from a vibrant community of students and enthusiasts, including olympiad champions, researchers, and professionals.

Synthetic Geometry Group-Ambuj Shrivastava's Proposal

Comments

Learn from a vibrant community of students and enthusiasts,
including olympiad champions, researchers, and professionals.