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Tom [10]
3 years ago
10

Given: circle k(O) with diameter AB and CD ⊥ AB Prove: AD·CB=AC·CD

Mathematics
1 answer:
Ksivusya [100]3 years ago
5 0

Answer:

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.

<em>(Refer the Attachment)</em>

<em>Now</em><em>,</em>

we get two triangles, as ADC and CDB

So

Side(AD)=side(BD)

CD is common side

and Angle(ADC)=angle(BDC)

<em>Using S-A-S test to triangles are similar </em>

<em>So their corresponding sides are in proportion </em>,

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.

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a) By distributive law of sets, we have;

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

From the complementary law of sets, we have;

A ∩ A' = ∅

Therefore, for A ∩ (A' ∪ B) = A ∩ B, we have

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