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# 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=$$2\times OC$$

Note by Ambuj Shrivastava
1 year, 10 months ago

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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, $$OA = 2 \times OC$$. · 1 year, 10 months ago

Nicely explained · 1 year, 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.)

$$3 (AB^2+BC^2+CA^2)= 4(AD^2+BE^2+CF^2)$$ · 1 year, 10 months ago

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 $$\dfrac {AN}{BL} \times \dfrac {BL}{BL} \times \dfrac {CM}{MA}=1$$ · 1 year, 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. · 1 year, 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 · 1 year, 10 months ago

Sorry I didn't know about that theorem so i thought it to be a good question but still the proof might be useful. · 1 year, 10 months ago

Yeah, It's okay. :) · 1 year, 10 months ago

Am i allowed to add more questions to the set because i recently found one. · 1 year, 10 months ago

Well, maybe you can add it as a different note for the benefit of everyone. · 1 year, 10 months ago

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. · 1 year, 10 months ago

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 · 1 year, 10 months ago

Yeah but I was thinking about its proof!! · 1 year, 10 months ago

You typoed the theorem. · 1 year, 10 months ago