2. Let D be that point on side BC of \(\triangle ABC\) such that AD is the bisector of \(\angle BAC\). Prove that \(\angle ADC\) is half the sum of the interior angle at B and the exterior angle at C. (I.e., show that \(\angle ADC = \frac{\beta + \epsilon}{2}\) in the figure.)
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We are given that AD is the bisector of ∠BAC. This means that ∠BAD is equal to ∠DAC. Show more…
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In Fig. $6.63, \mathrm{D}$ is a point on side $\mathrm{BC}$ of $\triangle \mathrm{ABC}$ such that $\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}}$. Prove that $\mathrm{AD}$ is the bisector of $\angle \mathrm{BAC}$.
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In triangle ABC, let AD be the angle bisector of the exterior angle at A, where D is the intersection of the exterior angle bisector with side BC (extended - so C - B - D). Prove that BD/DC = AB/AC. Hint: Draw a line through C that is parallel to AB to create similar triangles. Let BE be the interior angle bisector of angle B, where E is on AC, and CF be the exterior angle bisector of angle C, where F is on AB (extended - so B - A - F). Use what you proved in Theorem 2.4.15 (similar result for interior angle bisectors) to show that AF/FB * BD/DC * CE/EA = 1. Sketch and figure accurate enough to recognize that the lines AD, BE, CF are concurrent.
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AD is a median and an angle bisector of ΔABC. Show that ΔABC is isosceles.
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