+1452 1: Let line BC and line DE be chords of a circle, which intersect at A, as shown. If AB = 3, BC = 15, and DE = 3, then find AE.
2: In the adjoining figure, line AB is tangent at A to the circle with center O, point D is inside the circle, and line DB intersects the circle at C. If BC = DC = 3, OD = 2, and AB = 6, then find the radius of the circle.
3: In triangle ABC, AB = 10, AC = 8, and BC = 6. Let P be the point on the circumcircle of triangle ABC so that angle PCA = 45 degrees. Find CP.
4: Let P be a point outside of circle O. A segment is drawn from P such that it is tangent to circle O at point T. Meanwhile, a secant from P intersects O at A and B, such that PA < PB. If PA = 3 and PT = AB - PA, then what is PB?
Thanks so much!
+129928 1. We can find AE with the secant product theorem
CA * AB = AE * DE
18 * 3 = AE * AD
54 = AE * AD
Let AD = x
Then AE = 3 + x and we have that
54 = (3 + x)* x
54 = 3x + x^2 rearrange as
x^2 + 3x - 54 = 0 factor
(x + 9) (x - 6) = 0
Setting each factor to 0 and solving for x and we have that
x = -9 (reject) and x = 6 (accept)
So
x = AD = 6
And AE = 3 + x = 3 + 6 = 9
+129928 3. Note that triangle ABC is a 6-8-10 right triangle
And angle BCA = 90
And angle PCA = 45
So....angle BCP = 45
Then.....PC bisects angle BCA
Let the intersection of PC and AB = D
So.....in triangle ABC, since angle BCA is bisected, we have the following relationship
BC/CA = BD/AD
6/8 = BD/AD
3/4 = BD/AD
So....since BA = 10.....there are 7 equal parts of BA
BD = (3/7)*10 = 30/7
AD = (4/7)* 10 = 40/7
Note....sin BAC = 6/10 = 3/5
And sin PCA = sin 45 = 1/√2
So.....using the Law of Sines
AD/ sin PCA = CD/ sin BAC
(40/7)/ (1/√2) = CD / (3/5)
CD = (40/7)(3/5)*√2 = (120/35)√2 = (24/7)√2
And using the intersecting chord theorem
BD * AD = CD * PD
(30/7) (40/7) = 24√2 / 7 * PD
1200/7 = 24√2 * PD
PD = (1200) / ( 7 * 24 * √2)
PD = (25/7)√2
So....CP =
PD + CD =
(25/7)√2 + (24/7)√2 =
(49/7)√2 =
7√2
+129928 4. We can use the secant-tangent theorem to solve this, ACG...
We have something like this (not to scale) :
PA * PB = PT^2
PA * PB = (AB - PA)^2
PA * ( AB + PA) = AB^2 - 2PA*AB + PA^2
PA*AB + PA^2 = AB^2 - 2PA*AB + PA^2
3*AB + 3^2 = AB^2 - 2*3*AB + 3^2
3AB + 9 = AB^2 - 6AB + 9
3AB = AB^2 - 6AB rearrange as
AB^2 - 9AB = 0 factor
AB( AB - 9) = 0
Seting each factor to 0 and solving for AB we have that either
AB = 0 (reject) or AB = 9 (accept)
So.....PB = AB + PA
PB = 9 + 3
PB = 12
CPhill: Please check my reasoning and see where I went wrong. Thanks.
3)
Triangle ABC is a right-angle triangle. Angle BCA is 90 degrees. This angle is bisected from C to a point D on AB. The height of this line CD is 4.8. In triangle ABC, angle BCA =36.87, angle ABC=53.13.
In Triangle ACD, AD =8 / sin(45) = 5.657 and BD = 10 - 5.657 =4.343. But we have:
5.657 x 4.343 =4.8 x PD, and PD =5.11845. AD + PD =4.8 + 5.11845=9.91845 =~7Sqrt(2).
Note. But the answer isn't exactly 7sqrt(2) !!. Why? Can you spot the mistake?.
