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
