Geometry Problem

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

Click the figure below to see the complete problem 776.

## Saturday, June 30, 2012

### Problem 776: Triangle, Isogonal lines, Congruent angles, Perpendicular, Concyclic points, Circle, Center, Midpoint

Labels:
angle,
center,
circle,
concyclic,
congruence,
cyclic quadrilateral,
isogonal,
midpoint,
perpendicular,
triangle

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http://img585.imageshack.us/img585/2386/p776resuelto.png

ReplyDelete1) Prolong FO and HE to point P. Note that FD is paralell to HP. Then as O is midpoint of DE, it's easy too see that triangles FOD and POF are congruent (ASA). This implies that O is HPF hypotenuse's midpoint , i.e. O it's circumcircle. Analog for triangle GM, if we extend GD and MO to Q.

2) As <H + <M = 180°, and <F+<G=180°, the quadrilaterals BMEH and BFDG are cyclics. Then, looking the picture (2.a), (2.b) we can see that <FHM = <FGM. Hence FHGM is cyclic, and by (1) we have that O is its circumcircle.

Greetings go-solvers :)!

I would like to suggest another way to prove the concyclic part.

ReplyDeleteLet BD intersects EH at P and BE intersects DG at Q

By similar triangles,

(BH/BF) = (BP/BD) and (BG/BM) = (BQ/BE)

Say BP = kBH, BD = kBF, BQ = mBG, BE = mBM, where k,m are constants for ratios.

Now consider triangle BDQ ~ triangle BPE, hence

(BP/BE)=(BQ/BD)

=> BP*BD = BQ*BE

=> kBH * kBF = mBG * mBM

However, k = m as angleABD = angleEBC, so they have the same trigonometric ratio.

So BH * BF = BG * BM

By the converse of power of the circle at B, it is proved.

http://img402.imageshack.us/img402/5629/problem776.png

ReplyDeleteConnect lines per attached sketch

Note that quadrilaterals BFDG and BHEM are cyclic

We have ∠BFG=∠BDG and ∠BMH=∠BEH

We al so have ∆BHE similar to ∆ BGD…. ( case AA)

So ∠BDG=∠BEH ==> ∠BFG=∠BMH

So quadrilateral HFMG is cyclic.

Perpendicular bisectors of HF and GM intersect each other at O

So O is the center of qua. HFMG

The concylic part is very easily proved by the 2 sets of similar triangles without any additional construction yielding BF.BH = BG.BM which implies that FHGM is concylic

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