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# help geometry

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In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 60^\circ,$ $\angle ABC = 60^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$

Mar 25, 2024

#1
+394
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Because BAC = 60, ABC = 60, ACB should equal 60.

Therefore, ABC is equilateral.

Because AD is an angle bisector, then ABD, is 30, and so ADB = ADC = 90.

Therefore, set BD = x.

$${x}^{2}+{24}^{2}={(2x)}^{2}$$.

We get $$x = 8\sqrt{3}$$. (These triangle ratios are very special, see if you can find the ratios for a 30-60-90 triangle for yourself)

Therefore the area is $$\frac{8\sqrt{3}*24*2}{2}=192\sqrt{3}$$.

Mar 25, 2024

#1
+394
+2

Because BAC = 60, ABC = 60, ACB should equal 60.

Therefore, ABC is equilateral.

Because AD is an angle bisector, then ABD, is 30, and so ADB = ADC = 90.

Therefore, set BD = x.

$${x}^{2}+{24}^{2}={(2x)}^{2}$$.

We get $$x = 8\sqrt{3}$$. (These triangle ratios are very special, see if you can find the ratios for a 30-60-90 triangle for yourself)

Therefore the area is $$\frac{8\sqrt{3}*24*2}{2}=192\sqrt{3}$$.

hairyberry Mar 25, 2024