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

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

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## Sunday, September 11, 2016

### Geometry Problem 1259: Triangle, Circumcircle, Circle, Diameter, Parallel Lines

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BEFD is cyclic hence < CBD = < DEF

ReplyDeleteABCD is cyclic hence < CBD = < DAC

So < DEF = < DAC

Therefore EF //AC

Sumith Peiris

Moratuwa

Sri Lanka

https://goo.gl/photos/3nftCTs6WqmHt8pAA

ReplyDeleteConnect BD, BE and DF

Note that BE⊥AD and DF⊥BC

So Quadrilateral BEFD is cyclic => ∠ (BDE)= ∠ (BFE)= u

Qua. BACD is cyclic => ∠ (BDA)= ∠ (BCA)= u

So ∠ (BCA)= ∠ (BFE)= u => EF//AC

Problem 1259

ReplyDeleteLet K center of the circle with diameter AB and L center of the circle with diameter CD.

Let <BAD=x=<BCD and <ABE=90-x=<FDC .Then <EBF=<ABC-<ABE=<ADC-<FDC=<EDF.

So the EBDF is cyclic, then <FED=<FBD=<CBD=<CAD.Therefore EF//AC.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE