Question

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​

Answer 1

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

Learn more:

let o be any interior point of an equilateral triangle ABC .Show that 2 ...

https://brainly.in/question/11022661

find the incentre of Triangle formed by the vertices (0,0)(5,-1),(-2,3 ...

https://brainly.in/question/14969166

Prove that D is incenter of ABC

https://brainly.in/question/17356113