Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, November 1, 2014

### Geometry Problem 1053: Triangle, Two Perpendicular Medians, Midpoint, Congruence

Labels:
congruence,
median,
midpoint,
perpendicular,
triangle

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Obviously H is the centroid of ΔABC.

ReplyDeleteThus let M be the mid-point of AC, then BH=2×HM.

Now since ΔAHC is right-angled, thus M is the circumcenter.

Hence, AC=2×HM=BH.

Let F be the midpoint of AC

ReplyDeleteH divides AD as well as CE as 2:1.

AD = CE implies AH = CH and HE = HD.

So Right Δ’s AHE and CHD are congruent.

Thus AE = CD, so BA = BC implying BHF is the perp bisector of AC.

Hence ΔAHF is congruent to ΔCHF,

and each is right angled isosceles, HF = AF = FC,

Hence BH = 2HF = 2 AF = AC.

Fie F simetricul punctului H fata de punctul D=>BHCF paralelogram si AH=HF(deoarece 2HD=AH,H fiind centrul de greutate al triunghiului ABC)=>CH mediana si inaltime in triunghiul CAF=>CAE isoscel=>AC=CF

ReplyDeleteExtend BH to G such that H is mid point of BG. Then CH // AG by applying mid point theorem to Tr. ABG.

ReplyDeleteSimilarly AH//GC.

So AHCG is a //ogram with < AHC = 90, hence the same is a rectangle whose diagonals must be equal.

So AC = HG = BH

Sumith Peiris

Moratuwa

Sri Lanka