From an external point A, tangents are drawn to a circle, with the points of contact being B and C. D is mid point of the minor
arc BC. Prove that D is incenter of ABC
Answers
Given : From an external point A, tangents are drawn to a circle, with the points of contact being B and C. D is mid point of the minor arc BC
To Find : Prove that D is incenter of Δ ABC
Solution:
incenter of triangle where angle bisector meets :
Let say O is the center of circle
join BD & CD
D is mid point of BC
Hence
angle by arc BD center = angle by arc CD at center = angle by arc BD at center /2
=> ∠BOD = ∠COD = ∠BOC/2
hence in ΔBOD & ΔCOD
BO = CO = Radius
∠BOD = ∠COD
OD =OD common
=> ΔBOD ≅ ΔCOD
=> BD = CD
now in Δ ABD & ΔACD
AB = AC ( equal tangents)
AD = AD (common)
BD = CD (Shown above)
Hence Δ ABD ≅ ΔACD
=> ∠BAD = ∠CAD
=> AD is angle bisector
∠ABD = ∠ACD
Using The alternate segment theorem which states that in any circle, the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.
∠ABD = ∠BCD & ∠ACD = ∠CBD
∠ABD = ∠ACD hence
∠ABD = ∠CBD & ∠ACD = ∠BCD
=> BD is angle bisector of ∠ABC
& CD is angle bisector of ∠ACB
AD , BD & CD are angle bisector
Hence D is incenter of ΔABC
QED
Hence proved
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