please someone solve this problem.......
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Answer:
Hence proved...
AQ=1/2(BC+AB+CA)....
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Answer:
Step-by-step explanation:
Since tangents from an exterior point to a circle are equal in length.
Therefore, BP = BQ........[From B]... (1)
CP = CR.......[From C]... (2)
AQ = AR..... [From A]....(3)
From(3),we have
AQ = AR
=>. AB + BQ = AC + CR
=>. AB + BP = AC + CO.....
[From 1 and 2] ......(4)
Now,
Perimeter of triangle ABC=
BC + CA + AB..... (5)
Perimeter of triangle ABC=
AB + BC + AC
Therefore,
BC + CA + AB = AB + BC + AC
BC + CA + AB = AB +(BP + PC) + AC
BC + CA + AB=(AB + BP)+ (AC + PC)
BC + CA + AB = 2(AB + BP)
BC + CA + AB = 2(AB + BQ)
Therefore,
AQ = 1/2 (BC + CA + AB)
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