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 : * * =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=
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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 BLAN×BLBL×MACM=1
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Its complete proof would be as follows
triangle ALB similar to triangle CNB by AA criteria . .. ⇒ NBLB = CNAL → 1 triangle ACL similar to triangle BCM by AA criteria . ⇒ CLCM = ALBM → 2 triangle BMA similar to triangle CNA by AA criteria .. ⇒ MANA = BMCN → 3
Now, On multiplying 1,2 & 3.and rearranging we get
NBAN * LCBL * MACM=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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BLAN×BLBL×MACM=1 is true for any three cevians of a triangle, not only angle bisectors, or perpendiculars, or Medians
Ceva's theorem states thatLog in to reply
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Q2) From Appolonius' theorem, We have
AB2+AC2=2(AD2+4BD2)AB2+BC2=2(BE2+4AD2)AC2+BC2=2(CF2+4AB2)
Adding all these, and transposing (I'm sorry for posting the incomplete solution, It's night here.)
3(AB2+BC2+CA2)=4(AD2+BE2+CF2)
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Q3
We have ΔOAD similar to ΔOCE by the AAA criterion. (∠DOA=∠COE, ∠ODA=∠OEC, ∠OAD=∠OCE). Furthermore, 2CE=AD, which implies that ΔOAD is an enlargenment of ΔOCE by a factor of 2. This implies that OA=2OC since they are corresponding sides on the triangles. Therefore, OA=2×OC.
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Nicely explained
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