# Thread: triangle inscribed in the circle

1. ## triangle inscribed in the circle

triangle ABC is inscribed in the circle and AC = AB. The measure of angle BAC is 42 degrees and segment ED is tangent to the circle at point C. What is the measure of angle ACD?

2. Originally Posted by sri340
triangle ABC is inscribed in the circle and AC = AB. The measure of angle BAC is 42 degrees and segment ED is tangent to the circle at point C. What is the measure of angle ACD?
If you draw a circle, locate the centre, then the inscribed triangle is isosceles.
Join the centre to all 3 triangle vertices.
The angle at A is now split into 2 equal 21 degree angles.
If we label the centre F, then angle FAB = angle FAC = 21 degrees.
The 3 triangles within the inscribed triangle ABC are all isosceles also.
Hence, angle FCA = angle FBA = 21 degrees.

The tangent is perpendicular to the line CF.
Therefore the sum of angles FCA and ACD is 90 degrees.
Hence angle ACD is 90-21 = 69 degrees.

3. Hello, sri340!

Great solution, Archie!

Here's another approach . . .

Triangle $\displaystyle ABC$ is inscribed in the circle and $\displaystyle AC = AB.$

The measure of $\displaystyle \angle BAC$ is 42°
. . and segment $\displaystyle ED$ is tangent to the circle at point $\displaystyle C.$

What is the measure of $\displaystyle \angle ACD$ ?
Code:
A
* o *
*    / \    *  138°
*     /42°\     *
*     /     \     *
/       \         o D
*    /         \    *  /
*   /           \   * /
*  /             \  */
/ 69°           \ /
B o- - - - - - - - -o C
*               *
*           */
* * *   /
/
o E

Since $\displaystyle \Delta ABC$ is isosceles, $\displaystyle \angle B = \angle C = 69^o$

Then $\displaystyle \text{arc}(AC) = 138^o$
. .
An inscribed angle is measured by one-half its intercepted arc.

Therefore: .$\displaystyle \angle ACD \,=\,69^o$
. .
The angle between a common tangent and secant
. . . . is measured by one-half its intercepted arc.