## Saturday, May 9, 2020

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

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. 1. Solution sent by c.t.e.o :

1) 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

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

Let 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

3. < BLH = < BAH = < HEA = < BEL, so BE = BL
Similarly BM = BF = BE = BL hence MEFL is concyclic with B as centre

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

Sumith Peiris
Moratuwa
Sri Lanka