There is a ABCD square, inside him there is a P point. The angle BAP=angle ABP= 15° . Find the angle PCD.
Note: Do not use trigonometry.
Using the fact that triangles DPC and ABP are isoseles label all missing angles and set up a system of equations to solve.
We know that ABC = BAD = 75 and APB = 150.
Labelling APD = BPC = a and BCP = ADP = y and CDP = DCP = x and DPC = b
You should have 4 equations with four unknowns.
x+y = 90
2x+b = 180
2a+b+150=360
y+a+75=180
Now solve the system.
Hello Sorombo,
Did you try drawing the two circles.If you did you should see in addition to the two isosceles triangles ADP and CDE that the intersections of the circles meet along the perpendicular bisector of CD and AB. It follows that DCP is equilateral
Sorobam,
Choose a length of the side of square as a radius and draw the two circles.Connect the two circle intersections. This line is the perpendicular bisector of AB and DC. ADP is isosceles because each of its long sides is a radius but also DC, DP, and CP are radii so DPC is equilateral.