Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Saturday, August 19, 2017
Geometry Problem 1343: Triangle, Excircles, Circle, Tangent, Tangency Points, Chord, Perpendicular, 90 Degrees, Collinearity
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https://goo.gl/photos/he8TWHRyog7EizYN8
ReplyDeleteConnect AQ, AO, CQ, CO, OE and OD
Define points M and N per sketch
Let ED meet BO at H and FG meet ED at H’
Observe that AO, CQ, AQ and CO are angles bisectors of corresponding angles
Observe that AQ⊥AO, ED⊥AO, AQ⊥FG and AD⊥DO
1. Since AQ⊥AO and ED⊥AO => AQ //ED
Since FG ⊥ AQ so FG ⊥ED at H’ and MANH’ is a rectangle
2. Let u= ∠ (BAO)= ∠ (GFA)= ∠ (ODN)
And v= ∠ (BCQ)= ∠ (COD) ( see sketch)
Note that ∠ (AOD)= 90- u = ∠ (BOC)=> ∠ (BON)=v
We have triangle OCD similar to OHN… ( case AA)
So NH/CD= ON/OD => NH= (ON/OD). CD…. ( 1)
We have triangle FMA similar to DNO..( case AA)
So MA/FA= ON/OD = > MA=(ON/OD). FA= NH’… (2)
Note that CG=AD= p ( half of perimeter of triangle ABC)
So AG=AF=CD
Replace it in ( 2) we will have NH’= (ON/OD) . CD= NH
So H’ coincide to H’ and Q, B, H. O are collinear
Problem 1343
ReplyDelete1.Is AG=AF then <FGA=<BAC/2.So GF//AO, but AO is perpendicular at ED .So GF is perpendicular at ED.
2.Let GF cuts ED to H,then GH is perpendicular at HD. If the GF cuts QO at H’ then <QH’G=<EBH’-<BFH’=(<BAC/2+<ACB/2)-<GFA=<ACB/2=<QCA.So QH’CG is cyclic,then <QH’C=90=<QGC.But <CH’=90=<ODC,then CH’OD is cyclic,so <CH’D=<COD=<BCA/2=<QH’G.So GH’ is perpendicular at H’D.So the points H,H’
coincide.Therefore the points Q,B,H,O are collinear.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE
AG = AF so < AGF = A/2
ReplyDeleteAE = AD so < ADH = 90-A/2 considering Tr. AED
So in Tr. GHD, < G = A/2 and < D = 90-A/2 and it thus follows that < GHD = 90.