Geometry Problem. Post your solution in the comment box below.

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

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## Sunday, January 8, 2017

### Geometry Problem 1305 Triangle, Circumcircle, Angle Bisector, Arc, Perpendicular, Area

Labels:
angle bisector,
arc,
area,
circle,
circumcircle,
perpendicular,
triangle

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Problem 1305

ReplyDeleteLet AF, DG are perpendicular to BK, AB respectively.Then <AGD=<AHD=<AFD so the points A,G,D,F and H are concyclic .Hence <GFB=<GFD=<GAD=<BAE=<CAE=<CBE, so

GF//BE. Similar HF//EC.If the BC intersects GE and HE at K,L respectively .

Let area triangle ABC is (ABC) then (BGK)=(FKE) and ((HLC)=(FLE).

(ABC)=(AGKLH)+(BGK)+(HLC)=(AGKLH)+(FKE)+(FLE)=(AGEH)=2.(AEH).

APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

https://goo.gl/photos/2Wk2e5vJSCUmWPte8

ReplyDeleteDraw DF ⊥ AB

Triangle AHE congruent to AFE ( case SAS) ( see sketch)

FH⊥AE=> Area(AFEH)=AE.MH…= 2.Area(AHE)..(1)

Applying Ptolemy’s theorem in quad. ABEC give us AE= (b+c).BE/BC

Note that BE/BC= AD/(2.AH) and AD.HM=AH.DH ( relation in right triangle AHD )

Replace these values to (1) we will have Area(AFEH)=1/2. DH(b+c)

Now Area(ABC)=Area(ADB)+Area(ADC)= ½.DH(b+c)

Comparing these 2 areas we will have Area(AHE)= ½.Area(ABC)

Let DH = h, altitude of ∆AHD from H be p, AE = q and BE = CE = d.

ReplyDeleteFrom Ptolemy qa = d(b+c) ….(1)

S(ADC) = ½ hb = b/(b+c).S(ABC)…(2)

S(AHE) = ½ pq….(3)

p/h = ½ a/d …(4) from similar ∆s.

(3)/(2) => S(AHE) /S(ABC). (b+c)/b = pq/(hb) = ½ qa/(bd) from (4)

Now substituting for qa/d from (1),

S(AHE)/S(ABC) . (b+c)/b = ½ (b+c)/b

Hence S(AHE) = ½ S(ABC)

Sumith Peiris

Moratuwa

Sri Lanka

Join BE and form the triangle ABE

ReplyDeletem(AEB) = m(ACB) (angles in same segment) --------(1)

m(BAE) = m(EAC) = x (say) ----------(2) (given)

So triangles AEB and ACD are similar

=> AB/AE = AD/AC

=> AB.AC = AE.AD ----------(3)

Area of triangle AEH = 0.5*AE*AH*sin(x)

=> AEH = 0.5*AE*AD*cos(x)*sin(x)

=> AEH = 0.5*0.5*AE*AD*sin(2x)

=> AEH = 0.5*0.5*AB*AC*sin(2x) (from (3))

=> AEH = 0.5*(Area of ABC)

through AE, reflect HE,get H2E, H2 on AB;connect DH2;

ReplyDeletearea AHEH2 = 2x area AHE

connect HH2, AE perpendicular to HH2;

Through AO make diameter AK of circle O;

connect KH2, DK,HK,

also connect KB,KC,KE,

KB//DH2; KC//DH; KE//HH2;

area BDH2 =area DKH2, (KB//DH2); area DHC= area DHK, (KC//DH);

area KHH2 = area EHH2, (KE//HH2)

so area ABC =area AHKH2 =area AHEH2 =2x area AHE