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.
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