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# Geometry

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In triangle $ABC$, point $D$ is on side $\overline{AC}$ such that line segment $\overline{BD}$ bisects $\angle ABC$. If $\angle A = 45^\circ$, $\angle C = 45^\circ$, and $AC = 12$, then find the area of triangle $ABD$.

Jun 27, 2024

#1
+1230
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First, let's note that triangle ABC is a 45-45-90 triangle.

Thus, we can write

$$AB = BC = 12/\sqrt 2 = 6\sqrt 2$$

Also, let's note that from the problem, we have thart

$$AD = AD\\ BD = BD\\ AB = BC$$

This means that triangles ABD and CBD are congruent.

Thus, we have

$$[ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18$$

Thanks! :)

Jun 27, 2024

#1
+1230
+1

First, let's note that triangle ABC is a 45-45-90 triangle.

Thus, we can write

$$AB = BC = 12/\sqrt 2 = 6\sqrt 2$$

Also, let's note that from the problem, we have thart

$$AD = AD\\ BD = BD\\ AB = BC$$

This means that triangles ABD and CBD are congruent.

Thus, we have

$$[ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18$$