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Determinant

In a triangle...

Question

In a triangle ABC, D and E are points in BC such that $B A D = D A E = E A C = A / 3$ and $A D = 1$

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Solution

A) In $△ A B D \frac{sin B}{A D} = \frac{sin \frac{A}{3}}{B D} \Rightarrow B D = \frac{sin \frac{A}{3}}{sin B}$
B) $△ A D E \frac{sin \frac{A}{3}}{D E} = \frac{sin A E D}{A D} = \frac{sin (C + \frac{A}{3})}{A D}$
As $∠ A E D = C + \frac{A}{3}$ , gives
$D E = \frac{sin \frac{A}{3}}{sin (C + \frac{A}{3})}$
C) In $△ A E C \frac{sin \frac{A}{3}}{E C} = \frac{sin A E C}{A C} \Rightarrow E C = \frac{A C sin \frac{A}{3}}{sin (C + \frac{A}{3})}$ ...(1)
And in $△ A D C \frac{sin A D C}{A C} = \frac{sin C}{A D} \Rightarrow A C = \frac{sin (B + \frac{A}{3})}{sin C}$ (2)
From (1) and (2)
$E C = \frac{sin (B + \frac{A}{3}) sin \frac{A}{3}}{sin C}$
D) In $△ A D E \frac{sin A E D}{A D} = \frac{sin A D C}{A E} \Rightarrow \frac{sin (C + \frac{A}{3})}{A D} = \frac{sin (B + \frac{A}{3})}{A E}$
As $∠ A E D = C + \frac{A}{3}$ and $∠ A D C = B + \frac{A}{3}$ , gives
$A E = \frac{sin (B + \frac{A}{3})}{sin (C + \frac{A}{3})}$

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In a triangle ABC, D and E are points in BC such that BAD=DAE=EAC=A/3 and AD=1

In a triangle ABC, D and E are points in BC such that $B A D = D A E = E A C = A / 3$ and $A D = 1$