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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 :

Tushargupta0691idn

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

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