Question

i will insert a picture of the question, this is a 3 part question, i will inserted more information

Answer 1

Part A:

By Angle-Angle Theorem,

Triangle BCA is similar with triangle BDC (∠B ≅ ∠B, and ∠C ≅ ∠D)

Triangle CDA is similar with triangle BCA (∠CBA ≅ ∠DCA, and ∠BCA ≅ ∠CDA)

By transitive property

ΔBCA ~ ΔCDA ~ BDC

Part B:

$\begin{gathered} \text{By properties of similar triangle} \\ \frac{b}{c}=\frac{d}{b} \end{gathered}$

Part C:

Cross multiplying the proportions we have

$\begin{gathered} \frac{b}{c}=\frac{d}{b} \\ b\cdot b=cd \\ b^2=cd \end{gathered}$