Question

But, PA is a tangent and OA is the radius of the given circle.
-2 (180° - *
_PAB =
ZOAB+ZPAB = 90°
-
2x = { _APB
= ZOAB = 90° - (90°
-(90° - Ž{"
) = ZOAB
ZAPB = 220AB.
A
D
Q
podiple 14 In the given figure, the incircle of AABC
touches the sides BC, CA and AB at P, Q and
R respectively. Prove that
(AR + BP +CQ) = (AQ +BR+CP)
1
HH
B
P
SOLUTION
- 3 (perimeter of AABC).
We know that the lengths of tangents from an exterior point to
a circle are equal.
(i) [tangents from A]
AR = AO.​

Answer 1

Answer:

your answer is 40

Step-by-step explanation:

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