Well, here's a start!
A = angle A : so angle C = 100 - A which leaves angle BDC = 40 + A
a = BC, b = AC, c = AB, d = BD : so AD = a + d
Insert point E on AD such that DE = d, which leaves AE = a :
so we now have isosceles triangle BDE, equal sides = d, equal angles = (40 + A) / 2.
Insert point F on AB, such that EF is parallel to BC :
so we have triangle AFE, similar to triangle ABC
I think that should lead you to the solution: give it a shot...
Yes that was a faulty assumption.I calculated the value of AD,BD,DC and BC for a triangle where AB = 10.BC =3.947 DC =2.578, AD=7.422 BD=3.949
These values were calculated using trig. BC using cos 80/10 =1/2 of BC
using the angle bisector theorem BC/DC= 10/AD and from this 3.473 /DC = 10/10-DC
BD is found using the sin rule AC/sin60= BD/sin 80
Is there a geometry proof? I don't know
Something I don't understand...
Right, so if we knew that P, D and C are collinear, i.e., that P lies on AC, then the proof would be finished.P, D and C will be collinear only if
This is I am not sure about. I see how we find the angle QAD in the triangle QAD by subtracting angles Q and D from 180. However, it is not clear that since P may potentially not lie on AD.
That A, Q and B are collinear is known, but why AP = PQ is not clear (for the same reason as above).Therefore A, Q and B are collinear and triangle APQ is isosceles only if
I am not sure I understand this either. It seems that you assumed that and that the triangle ABC is isosceles to find its sides, whereas the problem was to prove that . In particular:Originally Posted by bjhopper(You must mean (cos 80) * 10 = 1/2 BC.) This assumes that the perpendicular from A divides BC is in half, which is not obvious. Also:cos 80/10 =1/2 of BCBy assuming that AD = 10 - DC, you assume that AC = AB.from this 3.473 /DC = 10/10-DC
I also came up with a trigonometry solution. Let . By the law of sines for BAD, . Also, for BDC, . Now, we may give BD any numeric value; then BC and will be determined uniquely because the whole picture can be scaled arbitrarily preserving . So, let BD = 1 and denote BC by a. From the first equation, ; from the second one, . Therefore,
It is easy to see that on (0, 100), the function is strictly decreasing: the denominator of the first fraction and the nominator of the second one increase while the denominator of the second fraction decreases. Therefore, there is at most one solution. With a bit of trig, it is possible to show that f(20) = 1.
Sorry, I spent a lot of time on this, and I am still confused. Could you write more precisely how and in which order you define various objects and what you know about them at that time, not what you would like to prove eventually?
We can have for free that the image of C is Q (so lies on AB) or that the image of B lies on AC. I thought you chose the first thing, making Q the image of C by definition. That's fine. But the rest of the proof uses the fact that the image of B (I presume, it is P by definition) is on AC, and I don't see it proved.At this point we don't know if it is possible for the rotated image of C to lie on AB
This we do know as |QD|=|DC| since triangles DBC and DQB have x, y and 40 degrees as shown.
Then we check the condition required for P to be on AC.