Given: ΔABC, AD bisects ∠BAC, AE ≅ ED To prove: AE/AC = BD/BC1 In ΔABC, AD bisects ∠BAC and AE ≅ ED ---- given 2 m∠BAD = m∠EAD ------- AD bisects ∠BAC 3 ΔAED is isoceles ------- AE ≅ ED, which is given 4 m∠EAD + m∠ADE + m∠AED = 180° ------ property of all triangles 5 m∠ADE = m∠EAD ------- property of isosceles ΔAED 6 m∠BAD + m∠EAD + m∠AED = 180° ------ substituting =s for =s (2 and 5) 7 ∠BAC = ∠BAD + ∠EAD ------- a whole is equal to the sum of its parts. 8. m∠BAC = m∠BAD + m∠EAD ------- measure of whole equals sum of measures of parts 9. m∠BAC + m∠AED = 180° -------- substituting =s for =s (6 and 8) 10. AB ∥ ED -------- if two interior angles on the same side of a transversal cutting two lines are supplementary then the lines are parallel" 11 AE/EC = BD/DC -------- If a line parallel to a side of a triangle intersects the other two sides, then it divides those sides proportionally. 12 AE/(EC+AE) = BD/(DC+BD) -------- Numerator addition property of a proportion. 13 AE/AC = BD/BC -------- AC = EC+AE and BD = DC+BD, substituting =s for =s Edwin