Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Saturday, May 9, 2020

### Dynamic Geometry 1479: Triangle, Circumcircle, Angle Bisector, Perpendicular Bisector, Chord, Concyclic Points, Parallel Lines, Step-by-step Illustration

Labels:
angle bisector,
chord,
circumcircle,
concyclic,
dynamic geometry,
GeoGebra,
IGS,
ipad,
ipadpro,
parallel,
perpendicular bisector,
step-by-step,
triangle

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Solution sent by c.t.e.o :

ReplyDelete1) Triangles EBM, BFL isosceles => B is center of the circle passes through M, E, F, L

2) Ang EFM = Ang HKM => EF//HK

Sketch for P1479

https://photos.app.goo.gl/cDnnRqmh2su1pcW99

https://photos.app.goo.gl/KQZZUh3UnhQKozyPA

ReplyDeleteLet u= ∠ (HAB)= ∠ (HEA)

And v=∠ (KCB)= ∠ (CFK)

And w=∠ (HKM)= ∠ (HLM)

1. quad. AHBL is cyclic => ∠ (HAB)= ∠ (BLH)= u= ∠(BEL)

EBL is isosceles and BE=BL

Similarly we also have BM=BF=BE

So M, E, F, G are concyclic with B as the center

2. In quad. MEFL we have ∠ (EFM)= ∠ (MLE)= w

In quad. HKLM we have ∠ (HKM)= ∠ (HLM)

So ∠ (EFM)= ∠ (HKM) => EF//HK

< BLH = < BAH = < HEA = < BEL, so BE = BL

ReplyDeleteSimilarly BM = BF = BE = BL hence MEFL is concyclic with B as centre

Therefore < MKH = < MLH = < MFE so EF // HK

Sumith Peiris

Moratuwa

Sri Lanka