Math  /  Geometry

QuestionA,BA, B and CC are points on a circle. - BCB C bisects angle ABQA B Q. - PBQP B Q is a tangent to the circle.
Angle CBQ=xC B Q=x Prove that AC=BCA C=B C

Studdy Solution

STEP 1

What is this asking? We're given a circle with a tangent line and some points, and we need to prove two line segments are the same length. Watch out! Don't mix up inscribed angles and central angles.
Also, remember the relationship between tangent lines and the radius to the tangent point.

STEP 2

1. Relate tangent-chord angles
2. Relate the bisected angle
3. Connect the angles
4. Conclude equality

STEP 3

Alright, let's **start** with the **alternate segment theorem**!
This theorem tells us that the angle between a tangent and a chord at the point of tangency is equal to the angle in the alternate segment.
In our case, this means that the angle CBQ\angle CBQ between the tangent PBQPBQ and the chord BCBC is equal to the angle BAC\angle BAC in the alternate segment.

STEP 4

So, we can write this as BAC=CBQ=x\angle BAC = \angle CBQ = x.
Remember this, it's going to be **super important** later!

STEP 5

We're told that BCBC bisects ABQ\angle ABQ.
What does bisect mean?
It means to cut something perfectly in half!
So, ABC\angle ABC and CBQ\angle CBQ are equal.

STEP 6

Since CBQ=x\angle CBQ = x, we know that ABC\angle ABC is *also* equal to xx.
So, ABC=x\angle ABC = x.
Another **key fact** to keep in mind!

STEP 7

Now, let's look at the triangle ABCABC.
We know two of its angles: BAC=x\angle BAC = x (from relating tangent-chord angles) and ABC=x\angle ABC = x (from relating the bisected angle).

STEP 8

Since the sum of the angles in a triangle is always 180180^\circ, we can find the third angle, ACB\angle ACB.
We have x+x+ACB=180x + x + \angle ACB = 180^\circ, which simplifies to 2x+ACB=1802x + \angle ACB = 180^\circ.
Therefore, ACB=1802x\angle ACB = 180^\circ - 2x.

STEP 9

Notice something **amazing**?
We have BAC=ABC=x\angle BAC = \angle ABC = x.
This means triangle ABCABC is an isosceles triangle!

STEP 10

In an isosceles triangle, the sides opposite the equal angles are also equal.
Since BAC=ABC\angle BAC = \angle ABC, the sides opposite these angles, BCBC and ACAC, must be equal.

STEP 11

Therefore, we've proven that AC=BCAC = BC!
Boom!

STEP 12

We have successfully proven that AC=BCAC = BC.

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