If AS is the bisector of angle BAC AB=10 AC=14 BC=6 BD=?
kvnmurty:
Is point D the intersection of AS and BC ?
Yes
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Let D be the intersection of BC & AS.
As AS bisects angle BAC, angle DAB = angle DAC = A/2
In Δ DAB, BD / sin A/2 = AD / sin B = AB / sin (B+A/2) --- eq 1
In Δ DAC, CD / sin A/2 = AD / sin C = AC / sin (B+A/2) -- eq 2
we used sin (180 - B - A/2) = sin (B+A/2)
Divide eq 1 by eq 2: to get :
BD / CD = Sin C / sin B = AB / AC
So BD/CD = 10/14 => add 1 on both sides, ( BD+CD) / CD = 24/14
BD + CD = BC = 6 => CD = 7/2 BD = 5/2
As AS bisects angle BAC, angle DAB = angle DAC = A/2
In Δ DAB, BD / sin A/2 = AD / sin B = AB / sin (B+A/2) --- eq 1
In Δ DAC, CD / sin A/2 = AD / sin C = AC / sin (B+A/2) -- eq 2
we used sin (180 - B - A/2) = sin (B+A/2)
Divide eq 1 by eq 2: to get :
BD / CD = Sin C / sin B = AB / AC
So BD/CD = 10/14 => add 1 on both sides, ( BD+CD) / CD = 24/14
BD + CD = BC = 6 => CD = 7/2 BD = 5/2
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