find the value of 3x

Geometry Level 2

$ABC$ is a triangle. Points $D$ and $E$ are on $BC$ and $AB$ respectively, such that $AD=4$ , $AE=6$ , $\angle ADE = \angle ACD$ , and $\angle BAD = \angle CAD$ . If the length of $AC=x$ , find $3x$ .

The answer is 8.

2 solutions

Elijah L
May 14, 2020

Observe that $\triangle ACD \sim \triangle ADE$ . Then:

$\displaystyle \frac{AE}{AD} = \frac{AD}{AC}$

$\displaystyle AC = \frac{AD^2}{AE}$

$\displaystyle AC = \frac{4^2}{6}$

$\displaystyle AC = \frac{8}{3}$

$3AC = \boxed{8}$

Nice using that approach is more easy to understand

Delbert McCullum - 1 year ago

Ron Gallagher
May 14, 2020

By the Law of Sines, sin(angle ADE) / 6 = sin (angle AED) / 4. This means sin(angle AED) = (2/3)*sin(angle ADE). Further, sin(angle ADC) / x = sin(angle ACD) / 4. Note that angle AED is congruent to angle ADC. Therefore, substitution yields:

(2/3)*sin(ADE) / x = sin (ADE) / 4, or

x = 8/3. Therefore, 3*x = 8

find the value of 3x

The answer is 8.

2 solutions

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