How cute is that triangle?

$\angle ABD = 180 - \angle ADB - \angle DAB = 180 - 90 - 70 = 20 ^\circ$ Extend AP such that it touches CB at Q. $\angle ABC = 50^\circ$ , so $\angle AQB = 180 - \angle QAB - \angle ABQ = 180 - 40 - 50 = 90 ^\circ$ , which implies that AQ is an altitude of triangle ABC.

Extend CP such that it touches AB at R. We know that the altitudes of a triangle are concurrent. Since AQ and BD are both altitudes of triangle ABC, then CR must be its third altitude.

Because they are vertical angles, $\angle CPD = \angle BPR = 90 - \angle PBR = 90 - 20 = 70 ^\circ$ . So, $\angle PCA = 90 - \angle CPD = 90 - 70 = 20 ^\circ$ .

In order to find the measure of $\angle PCA$ we will show that the triangles $ABD$ and $PCD$ are similar. It follows from the given angles that $\angle DAP = 30 ^ \circ$ and $\angle ABD = 20 ^ \circ$ . We now have $\angle DBC = \angle DAP = 30 ^ \circ$ and $\angle ADP = \angle BDC = 90 ^ \circ$ , which implies that the triangles $ADP$ and $BDC$ are similar. From the properties of similarity we get $\frac {AD}{BD} = \frac {PD}{CD}$ and therefore $\frac {AD}{PD} = \frac {BD}{CD}$ . We can observe that in triangles $ABD$ and $PCD$ the two matching sides have the same ratio in lengths, and the angles between these sides are equal in measure. Thus $\angle PCA = \angle ABD = 20 ^ \circ$ .

If we let $\angle PCA=\theta$ a straightforward angle chase gives that $\angle BCP=60-\theta,\angle ABD=20^{\circ},\angle DPB=30^{\circ},\angle BAP=40^{\circ},\angle PAC=30^{\circ}$ . From trigonometric form of Ceva's theorem, we get $$\frac{\sin 20^{\circ}}{\sin 30^{\circ}}\cdot\frac{\sin 30^{\circ}}{\sin 40^{\circ}}\cdot\frac{\sin(60-\theta)}{\sin\theta}=1$$ so $$\frac{\sin(60-\theta)}{\sin\theta}=\frac{\sin 40^{\circ}}{\sin 20 ^\circ}.$$ It is not hard to see that the only solution to this is $\boxed{\theta=20^{\circ}}$ . (this can be seen by writing each sin as a cosine and using product to sum identity).

Since $\angle{DBC}=30$ and $\angle{BDC}=90$ , $\angle{ACB}=60$ . Also, $\angle{PAC}=\angle{CAB}-\angle{PAB}=70-40=30$ .

Extend line $AP$ so that it intersects segment $BC$ at $E$ . Then $\angle{EAC}+\angle{ACE}=\angle{PAC}+\angle{ACB}=30+60=90$ , so $AE\perp BC$ . This means that $P$ is the intersection of altitudes $BD$ and $AE$ , so $P$ must be the orthocenter of $\triangle{ABC}$ . As a result, $\angle{PCA}+\angle{CAB}=90$ , so $\angle{PCA}=90-70=\boxed{20}$ degrees.

Let $AP$ intersect with $BC$ at $E$ . We have $\angle CAP = \angle CAB- \angle PAB=30^\circ$ . Thus, ABED is a cyclic quadrilateral because $\angle DAE = \angle DBE$ (i.e. $\angle CAP = \angle CBD =30^\circ$ ), and as a consequence we have $\angle AEB = \angle ADB =90^\circ$ . But this means $AE$ is orthogonal to $BC$ and $P$ is orthocenter of the triangle $ABC$ . Therefore, $\angle PCA = 180^\circ - 90^\circ - 70^\circ = 20^\circ$

Because $\angle BDC = 90 ^ \circ$ and $\angle DBC = 30 ^ \circ$ , we see that $\angle DCB = 60 ^ \circ$ and $\angle ABD = 20 ^ \circ$ .

$P$ is a point such that $\angle PAB = 40 ^ \circ$ . Since $\angle ABC = 50 ^ \circ$ , we see that the line through $A$ and $P$ is actually the perpendicular from $A$ onto $BC$ . Therefore, $P$ is the orthocentre of triangle $ABC$ .

As such the line through $C$ and $P$ is the perpendicular from $C$ onto $AB$ . Since $\angle BAC = 70 ^ \circ$ , we can conclude that $\angle PCA = 20 ^ \circ$

One can easily see than in right triangle ABC, angle ABC = 20.

Let unknown angle i.e. angle PCA =x Let, BC=a, using trigonometric ratios , in right triangle BDC, we can evaluate DC=a/2 and BD = a sqrt(3)/2

Now in right triangle ABD , AD = BD tan20= a (tan20) sqrt(3)/2 Now in right triangle APD , PD = AD tan30 = a (tan20) /2

Finally, in triangle PDC, tan x = PD/DC = (tan20)

Hence, x=20 degrees.

We are to find <PCA, and let's denote it as x. Given that <PAB= 40, we can easily determine <PAC = <CAB - < PAB= 70 - 40 = 30

Also, given that <DBC = 30, we can determine <ACB = 90- 30 =60, since triangle BCD is a right triangle.

How are we able to find x then? From the illustration made, we can notice that we will have 4 right triangles ( 2 smaller right triangles with common side PD and 2 bigger right triangles with common side BD )

Using the tangent function, we are able to interrelate the 4 right triangles.

So for triangle PAD, tan(30)= PD/AD =>1

for triangle PCD, tan(x)=PD/CD =>2

for triangle BAD, tan(70)=BD/AD =>3

for triangle BCD, tan(60)= BD/CD =>4

Equating PD from Eq. 1 and 2,

ADtan(30) = CDtan(x) => 5

Equating BD from Eq. 3 and 4,

ADtan(70) = CDtan(60)

CD = AD(tan(70)/tan(60)) =>6

Substituting CD from Eq. 6 to 5,

ADtan(30) = AD(tan(70)/tan(60))tan(x)

Thus, tan(30) = (tan(70)/tan(60))tan(x)

tan(x) = tan(30)tan(60)/tan(70)

We know that tan(30) = cot(60) and cot(60)tan(60) = 1

So, tan(x) = 1/tan(70) = cot(70) = tan(20)

so, x = 20 deg.

<PCA = 20 deg.

Since $D$ is a foot of a perpendicular, $\angle BDC = \angle BDA = 90^\circ$ . Since $\angle PAB = 40^\circ$ , $\angle PAC = 30^\circ$ , and since $\angle DBC = 30^\circ$ , $\angle DCB = 60^\circ$ and thus $\angle ABC = 50^\circ$ . Now let $E$ be the intersection of the lines $AP$ and $BC$ . Since $\angle PAB = 40^\circ$ and $\angle ABC = \angle ABE = 50^\circ$ , we have $\angle AEB = 90^\circ$ . We will now observe that $P$ is actually the orthocenter of the triangle $ABC$ . If we find the intersection $F$ of line $CP$ and $AB$ , we have a right triangle $AFC$ , and thus $\angle ACF = \angle ACP = 20^\circ$

We start by drawing a diagram and quickly angle chasing to find that <DAP=30, <PAB=40, <ABD=20 and <ACP+<PCB=60.

Note that \frac {DP}{DB} = \frac {\tan x}{\tan (x+y)}.

Let <ACP=x and <PCB=y, and let PD have length a. Then AD will have length a\sqrt{3}. Using the Law of Sines on triangle ABD, we see that \frac {a\sqrt{3}}{\sin20} = \frac {DB}{\sin70}. Then DB=\frac {a\sqrt{3}*\sin70}{\sin20} and we have \frac {a\sin20}{a\sqrt{3}\sin70} = \frac {\tanx}{\tan(x+y)}. Simplifying, and using the fact that tan(x+y)=60 as well as the fact that \sin70=\cos20, we find that tan 20 = tan x. The measure of x, then is 20 degrees.

Solution 1: Let $AD=1$ , then $DP = \tan 30^\circ$ , $DB = \tan 70^\circ$ and $DC = \tan 70^\circ \cdot \tan 30^\circ$ . Reflect $A$ about $BD$ to a point $A'$ . Since $DA' \cdot DC = DP \cdot DB$ , it follows that $A'CBP$ is a cyclic quadilaterial. Hence $40^\circ =\angle BAP = \angle BA'P = \angle BCP$ , which gives $\angle PCD = 180^ \circ - 70^\circ - 50^\circ - 40^\circ = 20^\circ$ .

Solution 2: Let $\angle PCD = \theta$ . By the sine-rule version of Ceva's theorem, we get that $\sin 40^\circ \cdot \sin 30^\circ \cdot \sin \theta = \sin 30^\circ \cdot \sin 20^\circ \cdot \sin (60^\circ - \theta)$ . Since $\theta = 20^\circ$ satisfies the equality, hence $\theta = 20 ^\circ$ .