Respuesta :

sqdancefan
12. ∆ABD is similar to ∆PQD, so
   QD/z = BD/x
Likewise ∆CDB is similar to ∆PQB, so
   QB/z = BD/y

Since QD + QB = BD, you have
   QD/z + QB/z = BD/x + BD/y
   BD/z = BD/x + BD/y
Dividing by BD gives the desired result:
   1/x + 1/y = 1/z


13. Triangles ABD, BCD, and ACB are all similar. This means
   AB/AC = AD/AB
   AB² = AC×AD = (4 cm)×(9 cm)
   AB² = 36 cm²
   AB = 6 cm
and
   BD/CD = AD/BD
   BD² = CD×AD = (5 cm)×(4 cm)
   BD² = 20 cm²
   BD = 2√5 cm

Otras preguntas