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Determine $\triangle ABC$, given the measure of $\angle A$, the length ($h_a$) of the altitude from $A$, and the length ($t_a$) of the angle bisector at $A$.

I found it in our geometry book. My problem in solving this is that I can't find a place to put $h_a$ in it. Then, with two points given, I could draw the third segment. How should I work?

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  • $\begingroup$ @bubba A is the measure of angle A.$h_a$ is the prepedicular drawn from angle A and $t_a$ is the bisector drawn from angle A. $\endgroup$ Nov 25, 2016 at 14:47

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I assume the original triangle to be named ABC, and name the angle at A as $\alpha$, to avoid confusion between the point A and the angle A.

If H is the point where $h_a$ meets the line through B and C and T the point where $t_a$ meets BC, then you can easily construct the triangle AHT, as you know 2 of its sides and the right angle at H. Now you know that $\sphericalangle BAT = \sphericalangle TAC = \frac{\alpha}{2}$. Since you constructed AT as part of the triangle AHT, you can now find the rays, starting from A, that contain B and C, resp. B and C can now be found by intersecting those rays with the line HT.

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