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△ABC是一等腰三角形，AB=AC，∠CAB=80°。P為△ABC內一點，使∠PBC=10°，∠PCB=30°，求∠PAC。

用旋轉那招??

try also

10° ^o^

We measured it is 10° , but no idea on it.

作一線AE使得AE=AB、角BAE=20及角CAE=60，連EB及EC。
由於AB=AE=AC及角CAE=60，故EC=AE=AC，即三角形ACE為正三角形。
由此得角BCE=60-50=10及角CBE=(180-20)/2-50=30；
故BP//EC and BE//PC，即BPCE為平行四邊形，亦得BP=EC=AC=AB；
故角BAP=(180-(50-10))/2=70；
得角PAC=80-角BAP=80-70=10。

marvellous!
I've approached this afternoon by constructing the parallelogram you've constructed, but didn't notice that's a equilateral tri.

有沒有圖?

I cannot use computer recently
I will post the detailed solution from yptsoi

May I?

of course, thx.

E在△ABC外，使AB=AE，∠BAE=20°，∠CAE=60°。連AE，BE及CE。

AB=AE (已知)
AB=AC (已知)
∴AE=AC
∴∠AEC=∠ACE (等腰三角形底角)
∠CAE=60° (已知)
∠AEC+∠ACE=180°-∠CAE (三角形內角和)
2∠AEC=180°-60°=120°
∴∠AEC=∠ACE=60°
∴△AEC是一等邊三角形
∴AC=CE=EA (等邊三角形性質)

∠ABC=∠ACB (等腰三角形底角)
∠CAB=80° (已知)
∠ABC+∠ACB=180°-∠CAB (三角形內角和)
2∠ABC=180°-80°=100°
∴∠ABC=∠ACB=50°
∠BCE=∠ACE-∠ACB
∴∠BCE=60°-50°=10°
∴BP//EC (錯角相等)

∠ABE=∠AEB (等腰三角形底角)
∠EAB=80° (已知)
∠ABE+∠AEB=180°-∠EAB (三角形內角和)
2∠ABE=180°-20°=160°
∴∠ABE=∠AEB=80°
∠CBE=∠ABE-∠ABC
∴∠CBE=80°-50°=30°
∴BE//PC (錯角相等)

∵BP//EC及BE//PC (已證)
∴BPCE是一平行四邊形 (根據定義)
∴BP=EC (平行四邊形對邊)
∴AB=AC=CE=EA=BP
∴∠BAP=∠BPA (等腰三角形底角)
∠ABP=∠ABC-∠PBC
∴∠ABP=50°-10°=40°
∠BAP+∠BPA=180°-∠ABP (三角形內角和)
2∠BAP=180°-40°-140°
∴∠BAP=∠BPA=70°
∠PAC=∠BAC-∠BAP
∴∠PAC=80°-70°=10°

解的真漂亮呀!!
真希望學會怎麼+輔助線

延伸CP到Q，使∠BAQ=∠CAQ=40°。連AQ，BQ及PQ。

∠CAB=80° (已知)
BA=CA (已知)
∠BAQ=∠CAQ=40° (已知)
AQ=AQ (公共邊)
∴△BAQ≡△CAQ (SAS)
∴∠ABC=∠ACB (等腰三角形底角)
∠ABC+∠ACB=180°-∠CAB (三角形內角和)
2∠ABC=180°-80°=100°
∴∠ABC=∠ACB=50°

∠BPQ=∠PBC+∠PCB (三角形外角)
∴∠BPQ=10°+30°=40°
∴∠BAQ=∠BPQ=40°
∠ACQ=∠ACB-∠PCB
∴∠ACQ=50°-30°=20°
∴∠ABQ=∠ACQ=20° (全等三角形的對應角)
∠QBP=∠ABC-∠PBC-∠ABQ
∴∠QBP=50°-10°-20°=20°
∴∠QBA=∠QBP=20°
QB=QB (公共邊)
∴△QBA≡△QBP (AAS)
∴BA=BP (全等三角形的對應邊)
∴∠BAP=∠BPA (等腰三角形底角)
∠ABP=∠ABC-∠PBC
∴∠ABP=50°-10°=40°
∠BAP+∠BPA=180°-∠ABP (三角形內角和)
2∠BAP=180°-40°-140°
∴∠BAP=∠BPA=70°
∠PAC=∠BAC-∠BAP
∴∠PAC=80°-70°=10°

This method is more beautiful.

yptsoi 寫到:This method is more beautiful.

both of them are nice

Someone tells me:

Reflect the triangle APC along AC to get triangle AP'C. Rotate the line P'C 60 degrees to get the line CP''. Finally reflect triangle AP'C along AP' to get triangle AP'P''. Now triangle CP'P'' is equilateral by cinstruction. Hence the angle PAC = 10 degrees.

I do not have time to read it, can anyone help me to translate?