Thread: Square and triangle.

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.

Here's a kicker,

BPC = APD = 90-15 = 75.

APB = 180 - (15+15) then you should be able to find CPD.

Can you please explain how you reason out 'BPC = APD = 90-15 = 75. '.
Thanks

hi pickslides

shouldn't PBC=75 and not BPC.

Thanks can you explain how the above facts allow x to be found.
Thanks

P is a point on the perpendicular bisector ofAB and DC
Altitude of APB =tan15*.5 whereAB=1
Altitude of DPC is 1-tan15*.5 =.866025
sin x = .866025 x=60

Originally Posted by pickslides

Here's a kicker,

BPC = APD = 90-15 = 75.

APB = 180 - (15+15) then you should be able to find CPD.

I don't understand why APD=90-15

could u explain it? i didn't get the argument

Hi guys, I meant PAD = 75

If you cannot use trig assume x=60 and work back showing that angles PAE and PEA equal 15.Any other angle for x will produce different ones

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.

There is another way. Draw two circles with a radius side of square fromC and D .ADP is then isosceles

Originally Posted by pickslides

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.

I think the system can't be solved. Because the 4th equation come from 1/2*(3rd-2nd)+1st

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

Originally Posted by bjhopper

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

I was thinking about the 2 circles, but i don't know how you know that triangle ADP is isosceles. In fact, I think I must show that point P belongs to the circle.

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.