Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

Given: circle k(O) with diameter AB and CD ⊥ AB

Prove: AD·CB=AC·CD

Given Circle KO With Diameter AB And CD AB Prove ADCBACCD class=

Sagot :

For a circle k(O) with diameter AB and CD ⊥ AB, it is proved that AD·CB=AC·CD, using similarity of triangles.

Given:

circle k(O) with diameter AB

CD ⊥ AB

To Prove: AD·CB=AC·CD

Proof:

In ΔADC and ΔCDB,

∠ADC = ∠CDB = 90°  

[∵Both are right angle triangles]

CB = CB [Common side]

⇒ AC / CB = CD / DB

Thus, ΔACD is similar to ΔCDB by RHS similarity.

Therefore, we can write,

AD/CD = AC/CB [Since corresponding sides of similar triangles are proportional]

⇒ AD·CB = AC·CD

Hence proved.

Learn more about similarity here:

https://brainly.com/question/25882965

#SPJ1

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.