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

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

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## Thursday, October 6, 2016

### Geometry Problem 1273 Circle, Tangent, Secant, Midpoint, Isogonal Lines, Congruent Angles

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https://goo.gl/photos/bYZ9Rdpg2VBXLXcB8

ReplyDeleteUsing pole and polar properties:

Let DE and DM cut circle O at F and G,

Let MA cut circle O at H

We have OM. OA= OC^2=OD^2

So A is the pole of BC w.r.t circle O

And A and E are harmonic conjugate points of D and F or ( A,E, D,F)= -1

Since ∠EMA= 90 so Apollonius circle diameter AE will pass through M

So ME is a angle bisector of angle MDF.

And ∠AMD=∠FMH=∠HMG => BC//GF

So Arc BG=Arc CF and ∠BDM=∠EDC

https://goo.gl/photos/bYZ9Rdpg2VBXLXcB8

ReplyDeleteUsing elementary geometry:

Let DE and DM cut circle O at F and G,

Let MA cut circle O at H

We have OM. OA= OC^2=OD^2

So OA/OF=OF/OM => triangle MOF similar to triangle FOA ( case SAS)

And FA/FO=FM/MO….. (1)

Similarly triangle MOD similar to triangle DOA ( case SAS)

And DA/DO=MD/MO ….. (2)

Divide (1) to (2) and note that OD=OF

We have FA/DA=FM/MD => MA is a external angle bisector of angle DMF

And ∠AMD=∠FMH=∠HMG => BC//GF

So Arc BG= Arc CF and angle BDM= angle CDE

Problem 1273

ReplyDeleteIf AD and DM cut circle O at K and L.Then AB^2=AM.AO=AC^2=AD.AK. So MOKD is cyclic.

OD^2=OB^2=OM.OA so triangle ADO is similar with triangle DMO.Therefore <OMD=<ODA

or <AMD=<ODK=<OMK. So <KMC=<CMD =<KMD/2=<KOD/2=<KBD, then <CMD=<CBD+<BDL=<KMC=<CBD+<CBK. So <BDL=<KBC=<KDC.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

Reading thro the 2nd proof of Peter, give below a slightly modified approach

ReplyDeleteLet AD meet the circle at P.

Let < MAP = p, < PAC = q, < ACD = < DBC = < DPC = r and < ODP = < OPD = s

Then OC^2 = OM. OA = OD^2 = OP^2

So < MDO = < MPA = p hence OMDP is concyclic

Therefore < AMD = s and so

< BDM = 90-s-r .....(1)

Now < DOC = 2r so < CDE = 90-r-s (since < ODP = s)

Therefore < BDM = < CDE = 90-s-r

Sumith Peiris

Moratuwa

Sri Lanka

How about the converse "if < BDM = < CDE then M is the mid point"?

ReplyDeleteThe symmedian passes through the intersection of the tangents to the circumcircle at the extremities of the opposite side, i.e. AD is symmedian of tr. DBC.

ReplyDeleteExtend AD to meet the circle O at X.Connect AMO and form the right triangle ABO

ReplyDeleteIn the triangle ABO

AB^2 = AM.AO

But AB^2 = AD.AX

Hence D,M,O and X are concyclic points.

Let Angle XDC = EDC = @ and ODX = $

=> Angle OXD = $ => OMD = 180-$ => AMD = $ => BMD = 90+$ -------------(1)

Join OC and consider the isosceles triangle ODC

We know Angle ODC = @+$ = OCD => Angle DOC = 180-2(@+$) => Angle DBC = DBM = DOC/2 = 90-(@+$)----------(2)

Now consider the triangle BDM and from (1) and (2)

Angle BDM = 180-(90+$)-(90-(@+$)) = @ = EDC

Let DM extended meet the circle at X and let ADE extended meet the circle at Y. Let AMO extended meet XY at Z.

ReplyDeleteOM.OA = OB^2 = OD^2 = OY^2

So < OAY = < MDO = < MYO hence OMDY is concyclic with OD = OY and so OM bisects < XMY.

Easily MOZ is perpendicular to XY and so BC//XY

Therefore BX = CY in isosceles trapezoid BCYX and the result follows

Sumith Peiris

Moratuwa

Sri Lanka