## Saturday, November 1, 2014

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

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

Click the diagram below to enlarge it. 1. Obviously H is the centroid of ΔABC.
Thus 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.

2. Let F be the midpoint of AC
H 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.

3. 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

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