Points A,B,C are given on circle Γ such that AB=BC. The tangents at A and at B intersect again at point D. The line CD intersects Γ again at E. The line AE intersects BD at F. If AD=120, what is the length of FB.
This one has been answered before, just before you stopped posting the remainder.
Use intersecting chord properties. The intersection could be inside or outside the circle.
AB and CD are chord which intersect at O (inside or outside the circle).
Then OA*OB=OC*OD
To find the length of FB, we need to analyze the given information and apply relevant geometric properties.
First, let's label the points and lengths in the diagram:
- Let O be the center of the circle Γ.
- Let x represent the length of AB = BC.
- Let y represent the length of AD.
We are given that AD = 120, so y = 120.
To find FB, we need to determine the length of BD, as FB forms a segment with BD.
Now, let's use the properties of intersecting tangents:
- The tangent at point A forms a right angle with the radius OA.
- The tangent at point B forms a right angle with the radius OB.
- Thus, ∠OAD and ∠OBE are right angles.
Since AB = BC, the triangle ABC is isosceles. Hence, ∠ABC is also congruent to ∠ACB.
Note that ∠OAC and ∠OCB both subtend arc AC. So, they are equal (by the inscribed angle theorem).
Since ∠OAD and ∠OBE are right angles, the quadrilateral AODB is cyclic (i.e., all its vertices lie on a common circle).
Now, observe the circle ODBC. Angle-chasing within this cyclic quadrilateral, we can determine that:
∠BDP = ∠BCO [corresponding angles (inscribed angles)]
= ∠OAC [equal angles (opposite angles subtending same arc)]
= ∠ODA [equal angles (∠OAD = ∠OAC, as both are right angles)]
Thus, ∠BDP = ∠ODA.
Since OADB is cyclic (as previously established), we have ∠ODA = ∠OBA, as they both subtend the same arc AO.
Hence, ∠BDP = ∠OBA.
This shows us that triangles BPF and BDO are similar (due to having two pairs of congruent angles and sharing a common side).
Therefore, we can establish the following proportion:
BD/FB = BO/FP.
Since BO is equal to the radius of circle Γ (since OB is a radius), we can rewrite the proportion as:
BD/FB = r/FP,
where r represents the radius of the circle.
Now, observe that triangles BCO and BDO are congruent (since BC = BD and OB = OB).
Thus, triangle BCF is isosceles, with ∠CBF congruent to ∠CFB.
Finally, note that triangles BAO and BFO are similar (due to having two pairs of congruent angles and sharing a common side).
Therefore, we can establish the following proportion:
BF/AF = BO/AO.
Since AO is equal to the radius of circle Γ (since OA is a radius), we can rewrite the proportion as:
BF/AF = r/AB.
However, let's rewrite AB in terms of x (as defined at the beginning):
AB = BC = x.
Hence, we have:
BF/AF = r/x.
Solving this proportion, we find that:
BF = AF * (r/x).
Now, we need to express AF in terms of known lengths.
Observe triangles AOD and AFO:
∠AOD = ∠AFB [equal angles (angles subtending same arc AO)]
∠ODA = ∠OFA [equal angles (∠OAD = ∠OFA, as both are right angles)]
Thus, triangles AOD and AFO are similar.
So, we can establish the following proportion:
AO/FO = OD/AF.
Since AO is equal to the radius of circle Γ (since OA is a radius), we can rewrite the proportion as:
r/FO = y/AF.
Replacing y with its known value, we have:
r/FO = 120/AF.
Solving for FO, we find that:
FO = (r * AF)/120.
Now, let's substitute the value of FO into our previous equation for BF:
BF = AF * (r/x) = AF * [(r * AF)/120]/x = (AF^2 * r)/(120 * x).
At this point, we need one more equation involving AF in order to solve for both AF and FB.
Consider the triangle AFO:
Using the Pythagorean theorem, we have:
AF^2 + FO^2 = AO^2,
AF^2 + [(r * AF)/120]^2 = r^2.
Now, we have a system of two equations:
AF^2 + [(r * AF)/120]^2 = r^2,
BF = (AF^2 * r)/(120 * x).
To solve for AF and FB, you can use algebraic methods (such as substitution or elimination) to solve this system of equations simultaneously.