Known, as shown in the figure: ab = AC, angle BAC = 80 degrees, P is a point in the triangle ABC, angle PBC = 10 degrees, angle ACP = 20 degrees, find the degree of angle BAP

Known, as shown in the figure: ab = AC, angle BAC = 80 degrees, P is a point in the triangle ABC, angle PBC = 10 degrees, angle ACP = 20 degrees, find the degree of angle BAP

60 degrees. Here's a simple process of proof,
Prove twice congruence, make auxiliary line, make the middle line of triangle ABC, cross BP to F, connect CF, then take cm = CF on AC, connect PM, prove that triangle MCO is equal to triangle FCP, then prove that triangle amp is equal to triangle AFP, get angle PAC = angle PAF = 1 / 2 angle BAF, and know that angle BAC is equal to 80 degrees, so angle BAP = 60 degrees

As shown in the figure, a = 36 °, DBC = 36 ° and C = 72 °. Calculate the degree of 1 and 2 respectively, and explain which isosceles triangles are in the figure

△ABC,∠A=36°,∠C=72°,
Then, from the sum of △ internal angles, it is obtained that ∠ ABC = 72 ° and 〈 AB = AC,
BD is the angular bisector of ∠ ABC,
∴∠DBC=∠DBA=36°,∴DB=DA,
Then, from the sum of △ internal angles, it is concluded that ∠ CDB = 72 °, ∧ BD = BC,
The ABC, ADB and BCD are isosceles