Measure of a Bisecting Angle

Let $\angle CPA=\angle BPC=x^\circ$ . Since $AP$ is tangent to circle $\Gamma$ , we have $\angle OAP=90^\circ$ . It follows that $\angle AOP=180^\circ-(90^\circ+2x^\circ)=90^\circ-2x^\circ.$ We know that $OB=OA$ so $\angle OBA=\angle OAB=\frac12\angle AOP=45^\circ-x^\circ.$ Therefore, $\angle PCA=180^\circ-[x^\circ+(90^\circ+45^\circ-x^\circ)]=45^\circ.$

Let $D$ be the point such that $BD$ is be a diameter of $\Gamma$ . Let $E, F$ be the points where $PC$ (extended) intersects $\Gamma$ , with $PF > PE$ . Let $G$ be the intersection of $CP$ and $DA$ .

We recall that the measure of an angle outside a circle that is formed by two secants (or a secant and tangent) is half the measure of the positive difference of the intercepted arcs. Since $\angle BPC = \angle CPA$ , we know that arc(BF) - arc(DE) = arc(FA) - arc(EA). Rearranging the equation will give us arc(BF) + arc(EA) = arc(FA) + arc(DE).

We now use the fact that angle determined by two chords of a circle is equal to the average of the intercepted arcs. Multiplying 1/2 to the both sides of the previous equation, we conclude that $\angle PCA = \angle CGA$ . Thus since $\angle BAD = 90^\circ$ , it follows that triangle CGA is 45-45-90 triangle with $\angle GCA = \angle PCA = 45^\circ$ .

Draw out the diagram as specified by the problem. Make sure to note that Point $B$ is on the opposite side of the circle, because as stated in the problem $PO < PB$ .

Because $AP$ is tangent to Circle $\Gamma$ , label $\angle OAP = 90 ^ \circ$ .

If we set $\angle OAB = x ^ \circ$ , Because $OA$ and $OB$ are both radii, $\bigtriangleup OAB$ is isosceles, and the angles opposite to equal sides are equal, so $\angle OBA = x ^ \circ$ . Thus, $\angle BOA = (180-2x) ^ \circ$ .

Because $\angle AOP$ is supplementary to $\angle BOA$ , $\angle POA = (180-(180-2x)) ^ \circ = 2x ^ \circ$ .

The sum of the angles in a triangle is $180 ^ \circ$ . With $\bigtriangleup OAP$ this means that $\angle OAP + \angle APO + \angle POA = 180 ^ \circ$ . We can rewrite this as $90 ^ \circ + \angle APO + 2x ^ \circ = 180 ^ \circ$ . Thus, $\angle APO = (90 - 2x) ^ \circ$ , and as an angle bisector $\angle CPA = (45-x) ^ \circ$ .

Finally, we deal with the last triangle, $\bigtriangleup CPA$ . We have $\angle CAP + \angle APC + \angle PCA = 180 ^ \circ$ . Solving, we get $(90+x) ^ \circ + (45-x) ^ \circ + \angle PCA = 180 ^ \circ$ , $\angle PCA = \boxed{45 ^ \circ}$ .

let PO intersect circle at D let angle ABD=x angle AOD=2x angle PAO=90 because PA is tangent so in triangle PAO PAO + POA + APO = 180 90 + 2x + APO=180 APO=90-2x 2APC=APO APC=45-x OA=OB so ABO=BAO=x PAB=90+x in triangle APC PAC + PCA + APC = 180 PCA + 45-x +90+x=180 PCA=45

Let $BO$ extended meet $Γ$ at point M. Let $PC$ meet $Γ$ at point $Q$ and $PC$ extended meet $Γ$ at point $S$ . Let the point of intersection of $AM$ and $PC$ be $Q$ . Now according to the question $\angle CPA= \angle BPC$ . Thus $arc BS - arc QM = arc SA - arc QA \implies arc BS + arc QA = arc SA + arc QM \implies \frac{arc BS + arc QA}{2} = \frac{arc SA + arc QM}{2} \implies \angle PCA = \angle CTA$ . But since $Γ$ is tangential to $PA$ and point $A$ , $\angle BAQ= 90^\circ$ . Thus $90^\circ + 2\angle PCA = 180^\circ \implies \angle PCA= 45^\circ$ .

Picture can be found here

In triangle OPA, angle OPA be 2 alpha. Angle AOP = 90-2 alpha since OA perpendicular to tangent PA. Angle ABO = 45-alpha = Angle OAB since isoscles triangle. In triangle PCA, angle PCA =180 - angle CPA - angle CAP =180 - alpha - (90+angle OAB(=45-alpha)) =45

Suppose, angle APB=θ; _ _ _ _ _ _ _ (1)

We have ∠PAO=90; (Tangent and Radius)

So,∠AOP=90-2θ; (Sum of all angles of triangle=180)

Therefore, ∠AOB=90+2θ; (Supplementary angles)

Hece,∠OAB=45-θ; (Where OA=OB,Therefore∠OAB=∠OBA)

Further,∠PAC=135-θ; _ _ _ _ _ _ _ (2)

From (1) & (2);we can get : ∠PCA=45

Assume that, the first intersection of $PO$ with the circle is $D$ . Such that, $PO > PD$ . We see that $\angle BAD = \frac{1}{2} . \angle BOD = \frac{1}{2}. 180^{\circ} = 90^{\circ}$ . And, also since $PA$ is tangent line, it is satisfy that $\angle PAO = 90^{\circ}$ .

Assume that, $\angle OAB = \alpha$ . We obtain, that $\angle OAD = 90^{\circ} - \alpha$ , and $\angle DAP = \alpha$ . And, see that since $r = OA = OB = r$ . It's satisfy that $\alpha = \angle OAB = \angle OBA$ . So,

Look at $\triangle ABP$ , we see that $\angle BAP = (\alpha + 90^{\circ} - \alpha + \alpha) = 90^{\circ} + \alpha$ . And, $\angle ABP = \alpha$ . So, $\angle BPA = 180^{\circ} - \angle PBA - \angle BAP = 180^{\circ} - \alpha - (90^{\circ} + \alpha) = 90 - 2 \alpha$ . So, we obatain that, $\angle APC = \frac{1}{2}. \angle BPA = \frac{1}{2}. (90^{\circ} - 2 \alpha) = 45^{\circ} - \alpha$ .

So, $\angle ACP = 180^{\circ} - (\angle CPA + \angle PAC) = 180^{\circ} - (45^{\circ} - \alpha + 90^{\circ} + \alpha) = 45^{\circ}$

So, we obtain that $\angle PCA = \boxed{45^{\circ}}$

Let $\angle APC = \angle DPC = x ^ \circ$ . (\angle PAO = 90 ^ \circ), so that $\angle POA = (90-2x) ^ \circ$ ; so $\angle PBA = 1/2\angle POA = (45-x) ^ \circ$ . So $\angle PCA = \angle DPC + \angle PBA\ = 45^ \circ$ , by exterior angle property.

Let angle APB = a , let angle PBA = b, let angle BAO = c, let angle PCA = x

Since the lines OB and OA are both the same length as the radius of the circle, therefore the angle OBA and OAB are equal.

Therefore, we get b=c

Since line PA is tangent to the circle, then a line drawn from its radius to point A must be perpendicular, therefore angle PAO is 90.

We know that the angles of a triangle add up to 180, therefore we can create a few equations.

a+b+c+90=180 a+2b=90 a=90-2b (1)

Also, since angle CPA bisects angle APB, then angle CPA = a/2

Therefore, from triangle APC, we can also set another equation.

a/2 + x + d + 90 = 180 a/2 + x + d =90 (2)

By substituting (1) into (2), we get,

(90-2b)/2 + x + b = 90 45 - b + x + b = 90 45 + x = 90 x=45

Therefore, angle PCA is equal to 45 degrees.

First, draw a simple picture of a circle and draw a line tangential to that circle. Mark the point where the line touches the circle as A. Take any point on that tangential line as P. Then draw PO and extend it to intersect the circle at B (whereas B is on the extension of PO). Draw line AO and AB. Mark point C and draw line PC. (Note that it is not important to make sure that PC exactly bisects ∠APB, this picture is just used to visualise this problem.)

From the picture, we can see (from triangle APO): ∠APO+∠POA+∠OAP = 180° Since AP is tangential to the circle, ∠OAP = 90°. So, ∠APO+∠POA = 90°; ∠POA = 90° - ∠APO

From triangle CPB, we can see: ∠CPB+∠PBC+∠BCP = 180°; ∠BCP = 180° - ∠CPB -∠PBC

Because line PC bisects ∠APB, ∠CPB = (1/2) ∠APB. Point O and B are both on the line PB, so ∠APB = ∠APO. So, ∠CPB = (1/2) ∠APO.

∠POA is a central angle, with ∠ABP as its corresponding inscribed angle, so ∠ABP = (1/2) ∠POA. Point A and C are both on the line AB, so: ∠PBC = ∠ABP = (1/2) ∠POA

So, we may rewrite equation ∠BCP = 180° - ∠CPB +∠PBC as: ∠BCP = 180° - (1/2) ∠APO + (1/2) ∠POA. (Remember that ∠POA = 90° - ∠APO)

∠BCP = 180° - (1/2) ∠APO + (1/2) (90° - ∠APO)

       = 180° - \frac {1}{2} ∠APO + \frac {1}{2} (90°) + \frac {1}{2} ∠APO

       = 180° - 45° = 135°

From the picture we can see that ∠BCP + ∠PCA = 180°

∠PCA = 180° - ∠BCP = 180° - 135° = 45°

Let $PB$ intersect the circle at $Q$ with $PQ<PO$ . Now, $\angle OAP = \angle QAB = 90^{\circ}$ . Hence, $\angle QAP = \angle OAB = \alpha$ $\text{say}$ . And let $\angle BPC = \angle CPA = \theta$ .

So, in $\triangle QAB$ , $\angle Q + \angle B = 90^{\circ}$

$\implies$ $2 \times (\alpha + \theta) = 90^{\circ}$ or $\alpha + \theta = 45^{\circ}$

Thus, $\angle PCA = \angle CPB + \angle CBP = \alpha + \theta = \boxed{45^{\circ}}$

Let $Z$ be the other end of $PA$ , $D$ the other point of intersection of $PO$ with $\Gamma$ , $\angle BAZ = \frac{\stackrel \frown{BA}}{2} = \alpha$ and $\angle PBA = \frac{\stackrel \frown{AD}}{2} = \beta$

We have that $\angle PAB = 180-\alpha$ and $\angle APB = \frac{\stackrel \frown{BA}-\stackrel \frown{AD}}{2} = \frac{2\alpha-2\beta}{2} = \alpha-\beta \implies \angle APC = \frac{\alpha-\beta}{2}$

So $\angle PCA = 180-(180-\alpha)-\frac{\alpha-\beta}{2} = \frac{2\alpha-(\alpha-\beta)}{2} = \frac{\alpha+\beta}{2} = \frac{\frac{\stackrel \frown{BA}}{2}+\frac{\stackrel \frown{AD}}{2}}{2} = \frac{\stackrel \frown{BA}+\stackrel \frown{AD}}{4}$

But, since $DB$ is the diameter of $\Gamma$ , $\stackrel \frown{BA}+\stackrel \frown{AD} = 180^\circ$ , so

$\angle PCA = \frac{180}{4} = \boxed{45^\circ}$

When we draw the digram as according to question angle APB will be a right angled triangle that is it is 90 .When PC will bisect it ,it will be equal to 45.

To start of, assign PA as x, the radius as y, and PO as y+z. (Note: z represents the distance from the point P to the circumference of the circle.) Since OA is tangential to PA, angle OAP is right. Now, by Pythagorean theorem, (y)^2 + (x)^2 = (y)^2 + (2yz) + (z)^2. Canceling out reveals that (x)^2 =2(yz) + y^2. Since the length of the sides can change, while the angles must be kept constant, we can plug in 10 for x, 2 for z, and 24 for y. (In order to make calculation easier, smaller numbers are necessary). To find angle APO,we find the inverse tangent of y/x, or 2.4 , which is 67.38013505. To find angle BOA, we use exterior angle theorem, and add the previous angle measure to obtain 157.3801351. We subtract 180 by this result to get the sum of the angle measures of the other 2 angles. Since OB and OA are congruent, the opposite angles are congruent, so dividing the previous result by 2 would get the measure of both of the angles. We obtain 11.30993247 as the measure. To find the measure of angle APC, we find the length of the segment opposite to it using the angle bisector theorem. One should get 20/3 as the result. Then, using inverse tangent functions, the angle should be 33.69006753. Adding all this information up (the measures of the 2 angles and the right angles), we get 135. To get the final result, we do 180 - 135= 45 degrees.