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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1259.
BEFD is cyclic hence < CBD = < DEFABCD is cyclic hence < CBD = < DAC So < DEF = < DAC Therefore EF //ACSumith PeirisMoratuwaSri Lanka
https://goo.gl/photos/3nftCTs6WqmHt8pAAConnect BD, BE and DFNote that BE⊥AD and DF⊥BCSo Quadrilateral BEFD is cyclic => ∠ (BDE)= ∠ (BFE)= uQua. BACD is cyclic => ∠ (BDA)= ∠ (BCA)= uSo ∠ (BCA)= ∠ (BFE)= u => EF//AC
Problem 1259Let 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