Answer:
The correct option is;
(3) ∠ADB and ∠BDC
Step-by-step explanation:
From the division axiom, we have;
(a+a)/a = 2·a/a = 2
The given parameters are;
m∠ADC = m∠ABC
[tex]\overline {BD}[/tex] bisects ∠ADC and ∠ABC
Therefore;
∠ADB ≅ ∠BDC
∠ABD ≅ ∠CBD
Therefore;
∠ADB/∠CBD = ∠BDC/∠ABD
Given that m∠ADC = m∠ABC and ∠CBD = 1/2 × m∠ABC = 1/2 × m∠ADC = ∠ABD, we are allowed to say, ∠ADB and ∠BDC are equal.
