Given: △ABC, D ∈ ray BC
C ∈ BD, AB=AC=BC
Prove: BD>DA>AB

BD>DA>AB

Step-by-step explanation:

Since ACB = 60, ACD = 120. Since ACD>ACB then DA is greater than AB. Then we can see that AC + DC has to be greater than DA because of triangle property theorem. That's the same as BC + CD by transitivity and BC + CD is BD by PWP. That means that BD > DA. Since we know that BD > DA, we can say, BD > DA > AB.

y + 11 = (1/4)(x + 16)

step-by-step explanation: