In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2 OC.

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Solution

In $△ B D A$ and $△ O C A$
$∠ B A D = ∠ O A D = 45^{o}$ (common)
$∠ B D A = ∠ O C A = 90^{o}$ (angle in semicircle)
Hence,
$△ B D A$ and $△ O C A$ are similar (by AA)
$\frac{B A}{O A} = \frac{B D}{O C} = \frac{A D}{A C}$ and $B A = 2 r, O A = r$
$\frac{B D}{O C} = \frac{2 r}{r} \Rightarrow B D = 2 O C$

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In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2 OC.

In △BDA and △OCA∠BAD=∠OAD=45o (common)∠BDA=∠OCA=90o (angle in semicircle)Hence,△BDA and △OCA are similar (by AA)BAOA=BDOC=ADAC and BA=2r,OA=rBDOC=2rr⇒BD=2OC