Some random point

Consider $\angle ACP = \angle DCA - \angle DCP = 45^\circ - 20^\circ = 25^\circ = \angle DAP$ , and consider a circle centered at $B$ passing through $A$ (and $C$ ). Suppose the continuation of segment $CP$ intersects $AD$ at $E$ . And suppose the circle intersects the line $CE$ at $P'$ . Therefore $\angle ABP' = 2*\angle ACP' = 50^\circ$ . Therefore $\angle BAP' = \frac{180^\circ - 50^\circ}{2} = 65^\circ$ , which implies $\angle DAP' = 25^\circ$ . Therefore $P'$ is the intersection of the line of angle $25^\circ$ from $AD$ with the line of angle $20^\circ$ from $CD$ , which is $P$ . So $P'=P$ .

Therefore $P$ is on the circumference of the circle, and hence $|BA|=|BP|$ , therefore $\angle APB = \angle PAB = 65^\circ$

Denote the length of the square be $a$ . We draw the circle $(B,a)$ , we easily calculate that $\angle APC = 135 ^\circ$ , so we have $\angle APC = 180 ^\circ - \dfrac{\angle ABC}{2}$ , we can conclude that P lies on the above circle.

From there it's easily to check that $\angle APB = \frac{180 ^ \circ-\angle ABP}{2}= \frac{180^ \circ-2\angle ACP}{2}=\frac{180^ \circ-2*25^\circ}{2}=65^\circ$ .

The solution is completed.

Let the circle, $w$ , centered at $B$ passing through $A$ and $C$ intersect line $AP$ at a point $P'$ . Then since $\angle{BAD}=\angle{BCD}=90$ , $DA$ and $DC$ are tangents to $w$ . Since $\angle{DAP'}=25$ , $\angle{ABP'}=2\angle{DAP'}=50$ . Furthermore, $\angle{BAP'}=90-25=65\implies\angle{AP'B}=65$ . Also, $\angle{P'BC}=90-50=40$ , and $BP'=BC$ (both radii of the circle), so $\angle{BP'C}=\angle{BCP'}=70\implies\angle{DCP'}=20$ .

So $P'$ is a point such that $\angle{DCP'}=20$ and $\angle{DAP'}=25$ . We claim that $P'=P$ . It suffices to show that Point $P$ is unique.

Denote $F(l_1,l_2)$ as the acute angle formed by lines $l_1$ and $l_2$ . Then $AP$ and $CP$ are the unique lines such that $F(AP,AD)=25$ and $F(CP,CD)=20$ and $AP$ and $CP$ pass through the interior of the square (for $P$ to exist). Thus, lines $AP$ and $CP$ are fixed. But $P$ is the intersection of $AP$ and $CP$ , so $P$ is fixed as well. Thus, $P$ is unique.

As a result, $P'=P$ , so $\angle{APB}=\angle{AP'B}=\boxed{65}$ .

With B as centre and BA (= BC) as radius, describe a circle Γ. The major arc AC of Γ subtends the reflex angle 270 deg at the center B. Looking at the quadrilateral ABCP, we observe that ∠APC = 360 deg - (∠ABC + ∠PCB + ∠PAB) = 360 deg - (90 deg + 70 deg + 65 deg) = 135 deg (Half of 270 deg, the reflex ∠ABC) . So P lies on the circle Γ. It follows that BP = BA (each is radius).

Hence ΔABP is isosceles and ∠BPA = ∠BAP = 65 deg.

Note: I am not yet familiar with Latex. So my solution might apear long. May I know how I can attach figures along with solutions of Geometry Problems ?.

Since ABCD is a square, angle PAB=65 and anglePCB=70.

angle APC=360-(Angle PAB+anglePCB+90) angle APC=360-(65+70+90) angle APC=135

Construct a circle with center B and passing through points A & C. Therefore, the measure of the major arc AC is 270. Since A&C lies on Circle B & angle APC is 135, angle APC is an inscribed angle and it follows that P is also a point on circle B.

Triangle APB is an Isosceles triangle, AB=BP.

angle BPA = Angle PAB angle BPA= 65.

first assume a square of side 'a' placed such that its vertex D is at origin\ ( for simplicity) and AD along y-axis and DC along x-axis such that coordinates of each vertex are

A : (0,a) B: (a,a) C : (a,0) D : (0,0)

now angle DCP = 20 ( in degrees as given in question)

so slope of line CP = tan(180-20) = -tan20

now angle DAP = 25

slope of line AP = tan(90+25) = -cot25

now for eqn of line CP

(y-0)/(x-a) = -tan20

y = (a-x)(tan20) ----(1)

for line AP

(y-a)/x =-cot25

x = (tan25)(a-y) --------(2)

as point p is common to both lines coordinate of the pt. can found by solving eqn (1) and (2)

after solving the coordinates come out to be

X = a(1-tan20)(tan25)/(1 - (tan25)(tan20)) ---(3)

Y = a(1-tan25)(tan20)/(1 - (tan25)(tan20)) ----(4)

now let angle APB = k and angle made by line BP with a line parallel to x-axis = h

tan(k) = tan (90+25 - h) -----(5) (draw diagram and check)

now,

slope of line BP = tan(h)

using eqn (3) and (4), slope of line BP is

(Y-a)/(X-a) = tan(h)

solving by using values os X and Y from eqn (3) and (4)

tan(h) = (1 - tan(20))/(1 - tan(25)) _ (6)

now from eqn (5)

tan(90+25 - h) = tan(k) = (tan(90+25) - tan(h))/( 1 +(tan(h))(tan(90+25)))

now putting tan (90+25) = -cot25 and tan(h) fron eqn (6)

k = 65 degree = angle APB which is the required angle

From $\angle DCP=20^{\circ}$ and $\angle DAP=25^{\circ}$ , we get $\angle BCP=70^{\circ}$ and $\angle BAP=65^{\circ}$ . It follows that $\angle APC=360^{\circ}-90^{\circ}-70^{\circ}-65^{\circ}=135^{\circ}$ .

I claim that $BP=BA=BC$ . The proof is as follows:

It is well known that the lengths of the sides of a triangle are in the same order as the measures of the angles opposite each side. More precisely, in any triangle $XYZ$ we have $YZ\le XZ\le XY$ if and only if $\angle X\le \angle Y\le \angle Z$ .

Now returning back to the problem, if $BP < BA=BC$ , then $\angle APB>\angle BAP=65^{\circ}$ and $\angle CPB>\angle BCP=70^{\circ}$ , implying that $\angle APC=\angle APB+\angle CPB>65^{\circ}+70^{\circ}=135^{\circ}$ , a contradiction. In the same way we can show that $BP>BA=BC$ leads to a contradiction as well (just replace every "<" sign with a ">" sign), so we conclude that $BP=BA=BC$ . Finally, this means triangle $ABP$ is isosceles, hence $\angle APB=\angle BAP=\boxed{65^{\circ}}$ .

Draw a circle with center on B and radius AB. This circle is tangent with AD and CD.

Let E be the intersection point between CP and AD and let F be the intersection point between AP and CD.

Also, let M be the intersection point between the circle and AF, and let N be the intersection point between the circle and CE.

$\angle DCE$ is a semi-inscribed angle with arc CN, then $\angle CBN = 40 ^ \circ$ because is a central angle.

In the same way, $\angle DAF$ is a semi-inscribed angle with arc AM, then $\angle ABM = 50 ^ \circ$ because is a central angle.

But, $\angle ABC = 90 ^ \circ$ then $\angle MBN = 0 ^ \circ$ and M, N and P are the same point. Therefore, $\angle APB = 65 ^ \circ$

Let (\ k(B, BC) ) be a circle, and $T$ a point on $k$ that lies on the smaller arc $AC$ .

The central angle corresponding to $\angle ATC$ is $\angle ABC = 270^\circ$ , so $\angle ATC = 135^\circ$ .

$\angle APC = \angle ATC = 135$ , so $P \in k$ .

$\triangle ABP$ is isosceles, so $\angle APB = \angle PAB = 65^\circ$

Let the side length of square $ABCD$ be $1$ . In the square, $\angle BAP$ and $\angle BCP$ are complementary to $25^\circ$ and $20^\circ$ respectively, so they are $65^\circ$ and $70^\circ$ respectively. In addition, the sum of the angles in a quadrilateral is $360^\circ$ , so $\angle APC = 360^\circ-90^\circ-65^\circ-70^\circ=135^\circ$ .

Now, consider the two triangles $\triangle APB$ and $\triangle CPB$ and let $\angle APB = x$ . By the law of sines on $\triangle APB$ , $\frac{BP}{\sin{65^\circ}}=\frac{1}{\sin{x}} \Leftrightarrow BP = \frac{\sin{65^\circ}}{\sin{x}}$

By the law of sines on $\triangle CPB$ , $\frac{BP}{\sin{70^\circ}}=\frac{1}{\sin(135^\circ-x)} \Leftrightarrow BP = \frac{\sin{70^\circ}}{\sin(135^\circ-x)}$

Setting the two expressions for $BP$ equal to each other, we have $\frac{\sin{65^\circ}}{\sin{x}}=\frac{\sin{70^\circ}}{\sin(135^\circ-x)} \Leftrightarrow 2\sin{70^\circ}\sin{x}=2\sin(135^\circ-x)\sin{65^\circ}$

Note that by the cosine angle sum identity (sum-to-product identity), $\cos(\alpha-\beta)-\cos(\alpha+\beta)=2\sin(\alpha)\sin(\beta)$

Applying to our equation, $2\sin{70^\circ}\sin{x}=\cos(70^\circ-x)-\cos(70^\circ+x)$ and $2\sin(135^\circ-x)\sin{65^\circ}=\cos(70^\circ-x)-\cos(200^\circ-x)$

Therefore, $\cos(70^\circ-x)-\cos(70^\circ+x)=\cos(70^\circ-x)-\cos(200^\circ-x)$ so $\cos(200^\circ-x)=\cos(70^\circ+x)$

The only solution with $0^\circ\le x < 180^\circ$ is if $200^\circ-x = 70^\circ+x \Leftrightarrow x = 65^\circ$ .

Perform an inversion centered at D with radius = AD, and let each point X map to X'. Then A = A', C = C'. Since ABCD are concyclic, AB'C are collinear. DCB = 90deg, thus DB'C = 90deg. As DAC is an isosceles right triangle, B' is the midpoint of AC. P' is a point outside triangle ACD such that AP'D = PAD = 25deg, CP'D = PCD = 20deg. Let w be the circumcircle of ACP'. Since AP'C = AP'D+CP'D = 45deg = DAC = DCA, both DA and DC are tangent to w. It is known that P'D is then the symmedian of triangle P'AC, thus AP'B = 20deg. BPA = BPD - APD = P'BD - P'AD = ADB + AP'B = 45deg + 20deg = 65deg.

I drew a square with sides=1 and then took point P as (x,y)

Then I got two equations for x and y in terms of tan(20) and tan(25)

After solving the equations I got x and y.

Now I cut the triangle APB into 2 triangles APW and BPW by cutting line AB at W where AW=(x) and BW=(1-x)

Now I evaluated the two angles APW and BPW (40 and 25) which add up to give APB=65

Consider the circumcircle of $CAP$ . Since $\angle CAP = 45^\circ - 25^\circ = 20^\circ = \angle DCP$ , it follows that $CD$ is tangential of the circumcircle. Hence, the circumcenter lies on $BC$ , which is perpendicular to $CD$ at $C$ . Also, the circumcenter lies on the perpendicular bisector of $AC$ , which is the line $BD$ . Thus, $B$ is the circumcenter of $APC$ .

This shows that $BA=BP=BC$ , so $BAP$ is an isosceles triangle, which gives that $\angle APB=\angle BAP = 45^\circ+20^\circ = 65 ^\circ$ .