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

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

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## Friday, September 9, 2016

### Geometry Problem 1257: Triangle, Altitude, Perpendicular, 90 Degree, Measurement

Labels:
altitude,
degree,
measurement,
perpendicular,
triangle

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Join FD, quad. ACDF is cyclic. Angle

ReplyDeleteGFD = C,

Also quad. BFEC is cyclic.Hence Angle AFE=Angle JFG= C.

Angle FGD = Angle FGJ = 90

By ASA Tr.FGD is congruent to Tr. FGJ, hence JG=GD

Quad GHDB is cyclic Angle GDH =90-A

And Angle GDB = 90-B, Angle HDB=90-A +90-B=C. We get Angle AGL=C

Since Angle AGL = Angle AFE, JP is parallel to GL, in Tr. JPD, G is midpoint of JD and GL is parallel to JP, hence JP=2×GL

JK + PK =2×GL

JK =2×10-9=11

https://goo.gl/photos/6syuJ3KQA1nGxwFk7

ReplyDeleteLet N is the orthocenter of triangle ABC

We have cyclic quadrilaterals AENF, ECDN, ACDF, BGHD and AEDB ( see sketch)

Let u= ∠ (EBC)= ∠ (DAC)= ∠ (HGD)= ∠ (FEN)

Let v= ∠ (BAD)= ∠ (BCF) and z=∠ (ABE)= ∠ (ACF)

We have ∠ (AFE)= ∠ (JFG)= 90-u => ∠ (FJG)= u => JE//GL

In right triangle ADC we have u+v+z= 90

In right triangle ADG we have ∠ (FDG)+v+z= 90 => ∠ (FDG)= u

Triangle FJD is isosceles => DG/DJ=DL/DP= ½

So JP= 2. GL= 20 => KJ= JP- KP= 11

Problem 1257

ReplyDeleteLet O is orthocenter the triangle ABC, the quadrilateral AFOE,EODC and AFDC are cyclic so

<FEO=<FAO=<FAD=<FCD=<OCD=<OED and triangles KHE ,DHE are equals(<EHK=<EHD=90)so

KH=DH.If EF intersects the circle passing through the points A,E,Dand B(<AEB=90=<ADB) in J’ then arc J’B=arcBD(J’B=BD , <KEH=<DEH=J’EB=<BED).But the diameter of the circle (J’,A,E,D,B) is AB so J’D is perpendicular bisector in AB.Then the points J,J’ coincide so JG=GD.Therefore GL//JP and PJ=2GL=2.10=20.But JK=JP-KP=20-9=11.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

We have a number of cyclic quads and accordingly < B = < GHK = < AEF = < EKD

ReplyDeleteHence GL // JP

But also < B = < DEC = < FEA hence Tr.s KHE and DHE are isoceles and KH = HD.

Hence from the midpoint theorem HL = 9/2 and so GH = 11/2 and therefore JK = 11

Sumith Peiris

Moratuwa

Sri Lanka

Correction

DeleteTr.s KHE and DHE are congruent not isoceles

A shorter cut would have been to realize that if O is the orthocentre of Tr. ABC then O is the incentre of Tr. DEF yielding both tr.s JFD and DEK isoceles.

ReplyDeleteEither result can then be used along with the mid point theorem to get the value of JK

Sumith Peiris

Moratuwa

Sri Lanka