## Friday, May 27, 2011

### Problem 606: Isosceles Triangles, Perpendicular and Parallel Lines

Geometry Problem
Click the figure below to see the complete problem 606.

#### 4 comments:

1. B, G, F, M, N, H are concyclic.
Let R be the radius of the circumscribing circle.
MG = 2R sin ∠MBG and HN = 2R sin ∠NBH
But ∠MBG = ∠NBH
So MG = HN subtend equal angles at N, G respectively
Hence GH ∥ MN

2. MG= HN is not enough to show that GH//MN
. please provide additional details to show that GH//MN

3. MGHN is a cyclic quadrilateral with the pair of
opposite sides MG and NH of equal length.
Triangles GHM and HGN are congruent by S.A.S property
(since MG = HN, GH is common and ∠GHM = ∠GBM = ∠NBH = ∠NGH)
So ∠MGH = ∠NHG
But ∠MGH and ∠MNH are supplementary
Hence ∠NHG and ∠MNH are supplementary which implies GH ∥ MN

4. Due to cyclic quadrilaterals made from right angles, (<GHF=<GBF)+(<FNM=<FBM)=<GBM
Because of isosceles triangles and cyclic condition of BHNF, <GBM=<NBH=<NFH
Left with <GHF+<FNM=<NFH from transitivity, if HF cuts MN at P then <FPN=<NFH-<FNM=<GHF and we are done