## Sunday, August 28, 2016

### Geometry Problem 1252: Triangle, Circle, Circumcircle, Altitude, Area, Hexagon, Perpendicular, 90 Degrees. Mind Map.

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1252.

1. Let H=Orthocenter

∠DBC=∠DAC=∠EBC
Similarly, ∠DCB=∠FCB
Thus, D and H are symmetric about BC.

Therefore, area of BDCH = 2× area of BCH.
By symmetry, the result follows.

2. This comment has been removed by the author.

3. Label the intersection of AB and FC as M.
Label the intersection of AC and BE as N.
Let the intersection of BE and FC be Q.
Then BMNC is a cyclic quadrilateral, hence angle MCN = angle MBN.
But angle FBA = angle FCA (= angle MCN) by considering the circumcircle.
Hence angle FBA = angle ABN.
Therefore triangle FBM is congruent to triangle QBM (by ASA) and has the same area.
Similarly for all such triangles around the edge of ABC, which means each small blue right-angled triangle inside ABC is mirrored by an orange one outside.
Thus the orange shading has the same area as the blue, and the hexagon area is double the area of ABC.

4. Problem 1252
Let BE, AD, CF intersects AC, BC AB at K, L and M respectively.Is <EAC=<EBC=<CAD(perpendicular line) and <ACE=<ABE=<FCA (perpendicular line).So triangle AHC=triangleAEC (H=orthocenter triangle ABC). Similar triangleAHB=triangle AFB, triangleBHC=triangle BDC.Area triangle ABC is (ABC) then
(AECDBF)=((AEC)+(AHC)+(CDB)+(CHB)+(BEA)+(BHA)=2.(AHC)+2.(CHB)+2.(BHA)=
2.(ABC).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

5. Can do without pen and paper

If H is the orthocentre of Tr. ABC, < ABF = < ABH and so the triangles ABF and ABH are congruent and so equal in area.

Similarly for BCF and BCH
Also for CAF and CAH