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

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

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## Wednesday, July 4, 2018

### Geometry Problem 1359: Triangle, Circumcircle, Congruent Angles, Midpoints

Labels:
angle,
circumcircle,
congruence,
geometry problem,
midpoint,
triangle

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https://photos.app.goo.gl/v33PQpQuJdPbRY1aA

ReplyDeleteConnect DA, DC, DB

Let angle(DBA)=angle(DCA)= u

Note that triangle (DFB) similar to triangle(DEC) … ( case AA)

So DE/DF= DC/DB and angle(EDC)=angle(FDB) , angle (EDF)= angle (CDB)

So triangle(EDF) similar to $(CDB) …(case AA)

And $(CDB) is the image of $(EDF) in the spiral similarity transformation with center D, angle od rotation = angle (EDC)

And dilation factor= DC/DE

Median DG is the image of median DH in this transformation so $(DEC) similar to $(DHG)

So angle (DHG)= angle (DEC)

Considering usual triangle notations, Join DC and DB

ReplyDeleteSince m(AED)=m(AFD) => ADFE are concyclic

=>m(EDF)=m(EAF)=m(CAB)=m(CDB)=m(A)

Also m(DEF)=m(DAF)=m(DAB)=m(DCB)

Therefore triangles DEF and DCB are similar

and m(EFD)=m(CBD) -----------(1)

Connect DH and DG, which form the medians of the two similar triangles DEF and DCB

=> DH/FE=DG/BC

=> DH/FH=DG/BG --------(2)

From (2) and (1) triangles DHF and DGB are similar

=> DF/DH=DB/DG

=> DF/DB=DH/DG -----------(3)

From (3) triangles DFB and DHG are similar

Hence m(DHG)=m(DFB)=m(DEC)