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.

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- September 18th 2011, 01:56 PMSoromboSquare 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. - September 18th 2011, 02:35 PMpickslidesRe: Square and triangle.
Here's a kicker,

BPC = APD = 90-15 = 75.

APB = 180 - (15+15) then you should be able to find CPD. - September 19th 2011, 01:34 AMkingmanRe: Square and triangle.
Can you please explain how you reason out 'BPC = APD = 90-15 = 75. '.

Thanks - September 19th 2011, 01:40 AManonimnystefyRe: Square and triangle.
hi pickslides

shouldn't PBC=75 and not BPC. - September 19th 2011, 02:11 AMkingmanRe: Square and triangle.
Thanks can you explain how the above facts allow x to be found.

Thanks - September 19th 2011, 11:59 AMbjhopperRe: Square and triangle.
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 - September 19th 2011, 12:43 PMSoromboRe: Square and triangle.
- September 19th 2011, 01:42 PMpickslidesRe: Square and triangle.
Hi guys, I meant PAD = 75

- September 19th 2011, 01:54 PMbjhopperRe: Square and triangle.
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

- September 19th 2011, 02:13 PMpickslidesRe: Square and triangle.
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. - September 19th 2011, 02:15 PMbjhopperRe: Square and triangle.
There is another way. Draw two circles with a radius side of square fromC and D .ADP is then isosceles

- September 19th 2011, 02:23 PMSoromboRe: Square and triangle.
- September 19th 2011, 04:10 PMbjhopperRe: Square and triangle.
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 - September 19th 2011, 04:20 PMSoromboRe: Square and triangle.
- September 19th 2011, 04:34 PMbjhopperRe: Square and triangle.
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.