Question

given below, P is a point of intersection of two circles with centres C and D. If the st. line APB is parallel to CD, prove that AB = 2CD.
AB​

Answer 1

Answer:

Here is you answer

Step-by-step explanation:

ANSWER

∵AB is parallel to CD

⇒CC

1

=DD

1

and hence, C

1

d

1

=CD→(a)

In △AC

1

C and PC^{1}C$$

⇒AC=PC=radius of I circle.

∠CC

1

A=∠CC

1

P=90

0

CC

1

=CC

1

(common)

∴AC

1

C≅PC

1

C by R.H.S.

⇒AC

1

=PC

1

→(1)

Similarly, PD

1

=BD

1

→(2)

∴AB=AC

1

+PC

1

+PD

1

+BD

1

Using (1) and (2)

⇒AB=2PC

1

+2PD

1

⇒AB=2(PC

1

+PD

1

)

⇒AB=2C

1

D

1

From (a) AB=2CD