Question

in the given figure, OD is perpendicular to chord AB of circle whose centre is O and BC is a diameter. Prove that CA=2 OD

Answer 1

OC=OB(RADII OF SAME CIRCLE)
AD=DB(PERPENDICULAR ON THE CHORD FROM THE CENTRE BISECTS THE CHORD)
SO,
O IS MID PT OF CB & D IS MID PT OF AB
BY MID PT THM,
OD||AC & OD=1/2AC
OR AC=2OD

Answer 2

Answer:

The proof is explained step wise below :

Step-by-step explanation:

For better understanding of the solution, see the attached figure of the problem :

Given : OD ⊥ AB

To prove : CA = 2 OD

Proof :

In ΔBOD and ΔBCA

∠ODB = ∠CAB ( each of 90°)

∠OBD = ∠CBA ( common angles are equal to each other)

So, by AA postulate of similarity of triangles, ΔBOD ~ ΔBCA

Now, sides of similar triangles are proportional to each other.

$\implies \frac{CA}{OD}=\frac{BC}{BO}\\\\\implies \frac{CA}{OD}=\frac{2BO}{BO} \text{ (BO is radius of circle and BC is diameter)}\\\\\implies \frac{CA}{OD}=2\\\\\implies CA=2\cdot OD$

Hence Proved.