## Saturday, November 14, 2015

### Geometry Problem 1165: Right Triangle, Circle, Center, Radius, Cathetus, Perpendicular, 90 Degrees, Perimeter, Metric Relations

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

Click the figure below to view more details of problem 1165.

1. http://s25.postimg.org/q5h2r9irj/Prob_1165.png
Let h= altitude from B of triangle ABC
We have EC= BC and BF= EC. Cos(C) = BM= h
So BGHF is a square
Draw a circle with center at B and radius= BF= BG= h
This circle will tangent to AC at M
Perimeter of EHD=EH+HD+ED= HF+HG= 2. h=6
So h=3 and FG= 3.sqrt(2)

2. Let I be the incenter of ABC, BK the altitude. Easily <EBD=45, BE_|_CI, BD_|_AI, thus BE is bisector of <ABK, BD the bisector of <CBK, triangles BFE and BKE are congruent, and so are BGD and BKD, hence EK=EF ( 1 ), KD=DG ( 2 )and BF=BK=BG and BFHG ( 3 ) is a square. From (1), (2) we get FH+GH=p(DEH)=6, from (3) FH=3 and FG=3sqrt{2}.

Best regards

3. A second way to prove BFHG was a square: by symmetry IE=IB=ID, I is the circumcenter of tr BED, thus <DIE=90, DIEH cyclic and <HID=<HED=<ACB ( 1 ). Since BIDC is cyclic, from (1) we infer B-I-H collinear, thus BFHG is a square.

Best regards

4. Let BS be an altitude of Tr. ABC = x

< CBE = < CEB = 45 + A/2 since < C = 90-A. So < ABE = 45-A/2 and so < BEF = 45+A/2

Hence Tr.s BEF and BEM are congruent ASA. So BF = x

Similarly we can show that BG = x so BGHC is a square of side x

Now perimeter of Tr. DEH = EH + EM + MD + DH = EH + EF + GD + DH = FH + HG = 2x

So x = 3 and FG = 3sqrt2

Sumith Peiris
Moratuwa
Sri Lanka

5. If FG meets BE at M and BD at N then EN and DM are altitudes of Tr. BDE and Tr.,BNE and BMD are right isoceles