I can't think of any other ways but Trig Ceva......

Actually, the best solution I've seen has nothing to do with trigonometry.

k2c901_1 wrote: Actually, the best solution I've seen has nothing to do with trigonometry. Can you show the construction?

I'll wait for a few days, then I'll post it if no one furnishes the construction then. (Actually the construction is not mine...)

Very very easy question...

I think I have got the construction of (1)

Click to reveal hidden text

Construct as I mentioned in my last post. Since $\triangle{BCD}$ is an equilateral triangle, so $DC=BC$, $\angle{BCD}=60^{o}$. We already have: $\angle{PCB}=10^{o}$ and $\angle{ACP}=20^{o}$, so $\angle{ACB}=30^{o}$. So we also have $\angle{DCA}=30^{o}$. From $AC=AC$, we shall have: $\triangle{DCA} \cong \triangle{BCA}$. Hence $\angle{CDA}= \angle{CBA} = 20^{o}$ Now, use the fact that $DB=DC$ and $PB=PC$, so we must have $DP$ bisect $\angle{BDC}$. Hence: $\angle{BDP}= \angle{CDP}=30^{o}$. Hence: $\angle{ADP}=10^{o}$, so $\angle{ADP}=\angle{ABP}=10^{o}$, which gives us A, P, B, D lie on the same circle. Hence: $\angle{BAP}=\angle{BDP}=30^{o}$

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Nice construction!

k2c901_1 wrote: Nice construction! Thanks. Can you show the construction you mentioned above? Also, I am pretty sure the answer of number two is also $30^{o}$ and I have found out the triangle in (2) is congruent to the triangle in (1). But I am kinda stuck here......

As soon as I get through my notes... I saw a solution using the excenter of a certain triangle, but I don't seem to remember it. Can anyone help me?

For (a): Choose $E$ on $AC$ so that $\angle EBC=30 ^\circ$. Notice that $A$ is the excenter of ? Hope this is enough. I don't want to be too obvious; I hope that you can find ? yourself.

for (b): I'm not going to say too much, but a continuation of the excenter argument in (a) shows that $P$ is the incenter of some triangle (not necessarily given in the problem!)

k2c901_1 wrote: For (a): Choose $E$ on $AC$ so that $\angle EBC=30 ^\circ$. Notice that $A$ is the excenter of ? Use the same method I posted above, we can prove that: $EP$ bisects $\angle{BEC}$. Hence: $\angle{BEP}=\angle{AEP}=\angle{AEF}=60^{o}$. Hence $A$ is the ex-centre of $\triangle{BEP}$. So $\angle{AEP}=40^{o}$. Which gives: $\angle{PAC}=100^{o}$. Since $\angle{BAC}=130^{o}$, so $\angle{BAP}=30^{o}$.

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Actually, the fact that $EP$ bisects $\angle BEC$ is trivial if you notice the symmetry between $B$ and $C$. (Note that $\angle EBC=\angle ECB$, and $\angle PBC=\angle PCB$.)

k2c901_1 wrote: for (b): I'm not going to say too much, but a continuation of the excenter argument in (a) shows that $P$ is the incenter of some triangle (not necessarily given in the problem!) Which one?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55281 \[{ \overline {\underline | \ How\ can\ I\ learn\ to\ draw\ on\ the\ computer,\ too\ ?\ | }} \]

Can anybody tell me which triangle in (b) is $P$ the incenter of?

Can anybody post a complete solution to part (b) of this problem?