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Mathematics, 20.06.2020 02:57 maddison788

In \triangle ABC, \bar{BD} is an altitude of \triangle ABC and also the bisector of \angle B. Prove that \triangle ABC is isosceles. This was the question I was given. This is my work: is an altitude of \triangle ABC. Therefore, it goes straight down from the vertex to be perpendicular with \bar{AC}. \bar{BD} is also a bisector of \bar{AC}. Therefore, it cuts \bar{AC} in half. Because \bar{BD} cuts \angle ABC in half by going straight down perpendicular to \bar {AC}, \bar{AB} and \bar{BC} are at the same angle to \bar{AC}. This means the triangle has two congruent angles, making it isosceles. (See Theorem 4.7 and Theorem 4.9).


\bar{BD}

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In , is an altitude of and also the bisector of . Prove that is isosceles. This was the question...

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