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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 975.
Let E be a point on AC such that DE⊥AC. Then ΔABD congruent to ΔAED (AAS). So AE=AB. Now ΔEDC is an isosceles right triangle. Thus EC=ED=BD. Hence, AC=AE+EC=AB+BD.
Point E is on AB extended such that EB=BD. Then <BED=<DCA=45, so triangles ADE and ADC are congruent. Therefore AC=AE=AB+BE=AB+BD
AC=V2 . AB ; CD=V2. BD (angle bisect law) BC=BD+CD= BD+V2.BD => BC+BD=2BD+V2.BDBC=BD+V2.BD =>AC=V2.BC=V2.BD+2BD => AC=BC+BD
Extend AB to E such that AE = ACThen Tr.s ABD and BCE are congruent ASA and so BD = BEHence AC = AE = AB + BE = AB + BDSumith PeirisMoratuwaSri Lanka