If a line drawn from one point of a triangle divides opposite side in ratio $1:2$ then in what ratio angle is divided by line?

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    $\begingroup$ In general, the information about the division ratio of the opposite side, by itself, does not give enough information to determine the division ratio of the angle. $\endgroup$ – André Nicolas Aug 3 '11 at 6:44
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    $\begingroup$ A simple illustration of the preceding comment: consider the median from the right angle of a right-angle triangle. It divides the hypotenuse in 1:1 ratio, but it divides the angle in the same ratio as the ratio of the two non-right angles, which can be any positive number. $\endgroup$ – Gerry Myerson Aug 3 '11 at 7:01
  • $\begingroup$ @Gerry: in fact generally it is not a ratio of the angles but a slightly more involved formula with sin() of the angles. $\endgroup$ – Jiri Kriz Aug 3 '11 at 12:52

Consider the triangle below with the usual notation. The division point D divides the edge $a$ in the ratio $r = a_1/a_2$.

enter image description here

Applying the law of sines to the triangle ABD we get

$$\frac{d} {\sin \beta} = \frac{a_1} {\sin \alpha_1}$$

And similarly for the triangle ADC we get

$$\frac{d} {\sin \gamma} = \frac{a_2} {\sin \alpha_2}$$

After dividing these two equations and some simple manipulation we obtain the ratio of the angles at $A$:

$$\frac{\sin \alpha_1} {\sin \alpha_2} = \frac{\sin \beta} {\sin \gamma} r$$

So, the ratio of the angles depends not only on the ratio $r$ of the opposite segments but also on the opposite angles as was already mentioned in the comment by @Gerry for a special case. The above equation gives an explicit general formula for this dependency.

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As an illustration: in the following the two thick black lines are two sides of two triangles, the thick red and blue lines (which are of equal length) are the third sides and the thin red and blue lines trisect the third sides. The different impacts on the angle at the top are dramatic. enter image description here

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