Your question can be quite confusing, but I think the gist of the question when paraphrased is: P<span>rove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?
Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector
Answer:
8/17
Step-by-step explanation:
right angle box means that area is the bottom
soh cah toa
hypotenuse is 17
hypotenuse is usually the biggest number
opposite is 8
sin is opposite/ hypotenuse
sin is 8/17
Answer:

Step-by-step explanation:
The zeros of a function is where the function crosses the x-axis. The x-intercepts are the zeros of a function.

Let output equal to 0.

Set factors equal to 0.


The zeros of the function are
and
.
Answer:
4(2x + 3) - x = 7x + 12
Explanation:
That's what T4L/Edge said