Question

i have exams tomorrow can you help, please!

Answer 1

Step-by-step explanation:

∆ ABC is formed on the circumference of semicircle.

Thus,  ∠CAB = 90°.

∠CAB + ∠CAD = 180° (linear pair)

90 + ∠CAD = 180°;

∠CAD = 90°

Since, DC is tangent on the circle.

Thus,  ∠BCD = 90 °.

Given,

BD is secant;

CD is tangent.

To prove : ∆ACD ~ ∆BCD;

Proof: In ∆ACD and ∆BCD;

 ∠CAD =  ∠BCD (each 90°)

 ∠D =  ∠D. (common)

Thus, ∆ACD ~ ∆BCD (by AA criteria of similarity)

That's all.