Find chord length given diameter and two other chords

Question

Problem:

I'm asked to find the length of $m$, given the following diagram.

Note that $\overline{AC}$ = 1 and $\overline{CD} = 1$ and that $\overline{AB}$ is a diameter whose length is 4.

Attempt:

I know that $\angle ADB$ is a right angle and that if I can find the length $\overline{AD}$ then this will be solved. I also know that $\triangle AGE \sim \triangle ADB$. And I now that the measure of $\angle ABD$ is half the measure of angle $\angle AED$ (one segment of which is not shown).

Question:

How can I go about finding the lengths of $\overline{AG}$ or $\overline{AD}$?

Answer 1

We know the following:

• $\triangle ADB$ is a right triangle (by the Inscribed Right Angle Theorem)
• $t \parallel k$ (by the Corresponding Angle Theorem)
• $\triangle AFE$ is a right triangle (by the Side Angle Side Similarity Theorem)
• $\triangle CFA$ is a right triangle (by the Vertical Angle Theorem)
• $\triangle CFD$ is a right triangle (by the Vertical Angle Theorem)
• $\triangle ADB \sim \triangle AFE$ (by the Side Angle Side Similarity Theorem)

We will use the Pythagorean Theorem and the following relationships to find $k$:

• $s + t = 2$ (since $s + t$ equals the radius of the semicircle)
• $s^2 + q^2 = 1^2$ (since $\triangle CFA$ is a right triangle)
• $q^2 + t^2 = 2^2$ (since $\triangle AFE$ is a right triangle)
• $2t = k$ (since $\triangle ADB \sim \triangle AFE$ by a factor of 2)

Based on the relationships above, we can form a system of equations to find the value of $k$:

\begin{align*} (2-t)^2 + q ^2 &= 1 && \text{substitution}\\ q^2 &= 1- (2-t)^2 \\ q &= \sqrt{1-(2-t)^2} \\ (\sqrt{1-(2-t)})^2 + t^2 &= 2^2 && \text{substitution}\\ -t^2 +4t -3 + t^2 = 4 \\ 4t-3 &= 4 \\ t &= \frac{7}{4} \\ 2(\tfrac{7}{4}) &= k && \text{substitution}\\ k &= \frac{14}{4} \\ k &= 3.5 \end{align*}

Thus, the measure of $k$ is $3.5$ units.

Answer 2

To find AG or AD draw the hight of triangle ACD to cross the side AD of triangle ABD at E.Also connect B to C. Triangle ACE is similar to triangle ABC for :

$\angle ABC=\angle CDA=arc AC/2$

So we have:

$\frac{CE}{AC}=\frac{AC}{AB}=\frac{1}{4}$ ⇒ $CE=\frac{1}{4}$

In right triangle ACE we have:

$AE^2=AC^2-CE^2=1^2-(\frac{1}{4})^2$ ⇒ $AE=\frac{\sqrt {15}}{4}$

⇒ $AD=2\times AE=\frac{\sqrt {15}}{2}$

If G is the projection of C on AB, then two triangles ACE and ACG are equal so we have :

$AG=CE=\frac{1}{4}$