如图,在直角三角形ABC中,角C=90度,AD平分角BAC,DE平行于CA,若CD=12,BD=15,求AE、BE的长

设DE=x∵DE//CA,∴∠ADE=∠CAD,又∠BAD=∠CAD,∴∠ADE=∠BAD,∴AE=DE=x∵DE//CA,∴△ABC∽△EBD∴BE/AB=BD/BC,∴BE/AE=BD/CD,而AE=x,BD=15,CD=12,∴BE=5x/4∵DE//CA,∴∠BDE=∠C=90°,∴BD^2+DE^2=BE^2即15^2+x^2=(5...