Network
m∠B = m∠D, m∠A = m∠A, m∠C = m∠B'
Maka ΔADB ∼ ΔABC (terbukti).
b. Tunjukkan bahwa ΔB'DC' dan ΔABC sebangun.
DC'/BA = C'B'/AC = B'D/CB
q/c = a/b = t/a
m∠B'DC' = m∠CBA, m∠DC'B' = m∠BAC, m∠C'B'D = m∠ACB
Maka ΔBDC ∼ ΔABC (terbukti).
6. Perhatikan gambar. (Lihat gambar di buku)
a. Tunjukkan bahwa ΔFCE ∼ ΔACB.
FC/AC = CE/CB = EF/BA
4/(4+DE) = 5/(5+10) = 6/(12+EF)
4/(4+8) = 5/(5+10) = 6/(12+6)
4/12 = 5/15 = 6/18
1/3 = 1/3 = 1/3
