Question

Ad and be are the altitudes of an isosceles triangle abc with ac=bc. prove that ae=bd

Answer 1

Solutions626.LetABCbe an isosceles triangle withAB=AC, and suppose thatDis a point on the sideBCwithBC>BD>DC. LetBEandCFbe diameters of the respective circumcircles of trianglesABDandADC, and letPbe the foot of the altitude fromAtoBC. Prove thatPD:AP=EF:BC.Solution 1.Since anglesBDEandCDFare both right,EandFboth lie on the perpendicular toBCthroughD. SinceABDEandADCFare concyclic,6AEF=6ABD=6ABC=6ACB=6ACD=6AFD=6AFE.Therefore trianglesAEFandABCare similar. ThusAEFis isosceles and its altitude throughAisperpendicular toDEFand parallel toBC, so that it is equal toPD. Therefore, from the similarity,PD:AP=EF:BC, as desired