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

Prove: AD·CB=AC·CD

Hence it is proved that AD.CB=AC.CD

Step-by-step explanation:

Given:

A circle with diameter AB and CD perpendicular to AB.

To Prove:

AD.CB=AC.CD

Solution:

Construct a circle with diameter AB and CD perpendicular AB

To get perpendicular Diameter  AB ,point C must be on Y-axis with passing through center getting CD as radius of circle.

(Refer the Attachment)

Now,

we get two triangles, as ADC and CDB

So

Side(AD)=side(BD)

CD is common side

and Angle(ADC)=angle(BDC)

Using S-A-S test to triangles are similar

So their corresponding sides are in proportion ,

AD/BD=AC/BC

therefore

AD*BC=AC*BD

as BD=CD ...........................  ( both are radius of the circle).

AD*BC=AC*CD

Hence it is required Proof.

x = 12

y = 15

step-by-step explanation: