jugoslav

If the ratio of BD:CD = 2 : 5    then                  10² - (2x)² = 17² - (5x)²

                                                                        x = 3          side  BC = 21

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In case of  3 : 5 ratio, we have:                    10² - (3x)² = 17² - (5x)²

                                                                                 x = 3.43693

                                                                       BC = 27.49544    Impossible!

Each side of quadrilateral ABCD is tangent to a circle. If AB = 84, BC = 78 and CD = 73, find DA.

w + x = 84         x = 84 - w

x + y = 78          y = 78 - x          y = 78 - (84 - w)         y = x - 6

y + z = 73          z = 73 - y          z = 73 - (x - 6)           z = 79 - x

AD = (79 - x) + x = 79

AB² = BC * BD               BC = 4.43908

AD = sqrt(AB² + BD²)

angle BDA = arctan(AB / BD)

r = (AD/2) / cos∠BDA

QP * PC = AP * PB

8 * sqrt(r² + 20²) = (r - 20)*(r + 20)

r = 25.70069473            r² = 660.5257097  

CP = 2/3 * [sqrt(3² + 4²)] = 10/3 

Triangle ABC, inscribed in a circle, has AB = 15 and BC = 35. A tangent to the circle is drawn at B, and a line through A parallel to this tangent intersects BC at D. Find DC. 

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Hello, Guest! I think that BC should be 25 instead of 35. If BC = 25 then DC = 16

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Let angle ABC = 90º

Angle BAD = arctan(15 / 35)

BD = tan∠BAD * AB = 6.428571429

DC = BC - BD = 28.57142857

AO = PO

PS = [sqrt(AS² + AO²)] - PO

PS = 1.973213749

Triangles ABC and ABD are isosceles with AB = AC = BD, and BD intersects AC at E.  If BD is perpendicular to AC, and angle C = 70 degrees, then find angle EAD.

angle EAD = 25º

Let midpoints of SR and RT be M and N

MAP and NBP are right angles

∠AMS = 63º              ∠BNT = 42º

AMNBP is a pentagon         (sum of interior angles of a pentagon is 540º) 

∠APB = 540 - (180 + 117 +138) = 105º

If the angle WXA = 22.5º then  AP = 5.991618724

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