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

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

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## Wednesday, October 5, 2016

### Geometry Problem 1272 Isosceles Triangle, Median, Midpoint, Perpendicular, 90 Degrees, Angles, Congruence

Labels:
90,
angle,
congruence,
isosceles,
median,
midpoint,
perpendicular,
triangle

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https://goo.gl/photos/bqyrjE7bstYM7kCDA

ReplyDeleteDraw rectangle AMNF ( see sketch)

Note that E is the midpoint of AB and FN.

∠NAM=∠MCN( ANC is isoceles)

And ∠NAM=∠FMA( AMNF os a rectangle)

So ∠FMA=∠MCN => FM//NC => ∠FMC=∠NDM= 90 degrees

In trapezoid FMDN , since MD ⊥ FM and E is the midpoint of NF so EM=ED

E is the center of cyclic quadrilateral AMDB => ∠BAD=∠BMD= ∠MCN

Problem 1272

ReplyDeleteIs <AMD=<CNB (they have their sides perpendicular).But AM/MD=MC/MD=NC/NM=NC/NB (AM=MC, NM=NB, triangle MCD is similar with triangle NCM).Therefore triangle AMD is similar with triangle CNB .So <MAD=<NCB.Therefore <DAB=<MCN (<BAC=<BCA ).

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

Area of Tr. MNC = ½ MD.NC = ½ MN.MC…(1)

ReplyDeleteBut AN = NC, NB =MN and AM = MC

So from (1), MD.AN = NB.AM

Hence in Tr.s ANB and AMD,

AN/NB = AM/MD.

Further the included angles

< ANB = < AMD

(since < ANM = < CNM = < CMD)

So it follows that the Tr.s ANB and AMD are similar and therefore

< ABM = < ADM.

Hence AMDB is concyclic and so

< BAD = < BMD = < CAN.

Sumith Peiris

Moratuwa

Sri Lanka

Peter's comment - "In trapezoid FMDN , since MD ⊥ FM and E is the midpoint of NF so EM=ED"

ReplyDeleteThis is so since the perpendicular bisector of MD passes thro' E

Sumith

ReplyDeleteThanks for your comment.

Peter

It is known that the perpendicular from N to AD passes through the midpoint P of MD. Indeed, let E be the projection of A onto CN, <PNE=<DAE (1), but triangles MND and CAE are similar, from (1) we infer P and D are homologous points, i.e. MP=PD, making NP||BD (*). With (1) we also get <MNP=<MAD (2). From tr. MNC: MN^2=ND.CN (3), but MN=BN, so from (3) we infer BN tangent to circle (BDC), or <NBD=<BCN; with (2) and (*) we conclude <CAD=<BCN, so we are done.

ReplyDeleteTr. CMD is similar to MND

ReplyDeleteCM/CD=MN/MD

=> 2.CM/CD=2.MN/MD

=> CA/CD=MB/MD

=> Tr. CAD is similar to MBD

=> m(DAC)=m(DBM)=> A,B,D,M are con-cyclic => m(BAD)=m(BMD)=m(DCM)