I don't remember my math proofing that well. I can only get as far as determining that ADFE is a rectangular parallelogram (not yet proven a square) because the angle of BDF and the angle of FEC are both 90-degrees and that means their supplementary angles are both 90-degrees as well. (The angle of DFE must also be 90-degrees.) I get lost there. I'm fairly certain I need to prove that BAC is a 45-90-45 right triangle so that I can prove that it's bisection (AF) makes AFB and AFC both 45-90-45 triangles too; but, I can't determine how to do that.
Edit: I got it. The punchline is, "I never said BAC is a triangle."
FJH said:I'm fairly certain I need to prove that BAC is a 45-90-45 right triangle so that I can prove that it's bisection (AF) makes AFB and AFC both 45-90-45 triangles too; but, I can't determine how to do that.
BAC does not have to form a right-angled isosceles. ADFE does not have to be a square. The bisection is not required to be perpendicular. It is drawn that way, but pushing point A off center will still allow the resultant figures to hold their stated properties (isosceles/rectangle).
If we do this the usual way, skipping a step or two in the process: AB and FE are parallel lines intersecting BC, meaning that ∠DBF=∠EFC. FEC and BDF are at least proportionate, but with their hypotenuse the same, we know they are congruent. AF splits the rectangle into two congruent triangles, with DF=AE and DA=EF. AE is congruent to CE. AFE and CFE share FE, so we say they are congruent. AF is then picking the hypotenuse of the triangle congruent to CFE.
I prefer a different sort of argument, excluding the degenerate solution of A=(B or C). The three line segments would form three separate radii of a circle. With A fixed at the end of a radius line, sweep A over the arc of the circle, while attached to the straight segments AB and AC. Notice that not only is ∠BAC fixed at 90 degrees over the arc, but that tan(∠ACB) is equal to |AB|/|AC| over the sweep.
Or more simply, as long as BC is a straight line, then the triangles AFB and AFC are mirror image isosceles right triangles (because, to make ∠BAC a right angle, and still have ∠BDF and ∠FEC be right angles, ∠BAF and ∠CAF must be 45 degrees, each), so BF = CF because of the symmetry, and AF = BF and CF because of the isosceles nature of the triangles.
Or more simply, as long as BC is a straight line, then the triangles AFB and AFC are mirror image isosceles right triangles (because, to make ∠BAC a right angle, and still have ∠BDF and ∠FEC be right angles, ∠BAF and ∠CAF must be 45 degrees, each), so BF = CF because of the symmetry, and AF = BF and CF because of the isosceles nature of the triangles.
Placing AF perpendicular to BC is a greatly simplifying assumption, but not justified by the givens. A counterexample:
Presume triangle AFC is equilateral and triangle AFB is isosceles, such that the angle of ∠AFB is 120 degrees (with ∠DAF = ∠DBF = 30 degrees). We can divide AFB with DF so that ∠DFB = ∠DFA = 60 degrees. DBF and ADF are 30-60-90 triangles.
Unless it's been lost in translation, you have to prove WHY AF=BF=CF, not verify IF AF=BF=CF. It's a given the three segments are equal. You just need to point out right triangle properties and the sum of all triangle angles being 180 degrees.
EDIT: Also note the hatch marks on segments BF and CF, that means both segments are of equal length. Yet more proof both angles at vertex A are 45 degrees and both sum a right angle.
Placing AF perpendicular to BC is a greatly simplifying assumption, but not justified by the givens. A counterexample:
Presume triangle AFC is equilateral and triangle AFB is isosceles, such that the angle of ∠AFB is 120 degrees (with ∠DAF = ∠DBF = 30 degrees). We can divide AFB with DF so that ∠DFB = ∠DFA = 60 degrees. DBF and ADF are 30-60-90 triangles.
∠DAF + ∠FAC = 30 + 60
If AFC were equilateral, then you couldn't fit a ∠AEF that was a right triangle in. (And, to use Asashio's logic, why would Kashima go out of her way to declare those points and make the dotted lines with marks for right angles if they weren't supposed to be used to find the answer?)
Since we already know that three of the angles of AEFD are 90 degrees, then AEFD is a square, because the sum of the angles have to add up to 360. So long as AB, AC, and BC are all straight lines with D, E, and F being points along those lines, then we can be absolutely certain of the exact angles, with ∠AFB being a right angle, and ∠ABF and ∠BAF being 45 degree angles.
NWSiaCB said: If AFC were equilateral, then you couldn't fit a ∠AEF that was a right triangle in.
You also can't have an equilateral right triangle unless the plane is curved.
--then AEFD is a square--
I don't think we've gotten that far yet. We can move A so that ∠BAC and ∠ADF and ∠AEF all stay as right angles but ADFE is stretched into a long-sided rectangle. AB does not necessarily have to be equal in length to AC. We can draw many perpendicular lines extending inwards from AB and AC that will intercept BC at the same point (and one of those points will always be F). If we do shift A around, however, AF is no longer the height - it will become a hypotenuse of the height segment perpendicular to BC.
I think the trick here is application of the altitude rule of triangle theorems. The height of a right triangle is a mean proportional between the segments of the hypotenuse that it divides. In the case of a 45-90-45, that could be used as proof for AF==BF==CF, since AF would be the altitude of BAC in that case, perpendicular to BC, and BC is perfectly divided in half by it. That works in the ideal circumstance but can we extend it for a right triangle where A isn't directly above the midpoint of BC?
You also can't have an equilateral right triangle unless the plane is curved.
I don't think we've gotten that far yet. We can move A so that ∠BAC and ∠ADF and ∠AEF all stay as right angles but ADFE is stretched into a long-sided rectangle. AB does not necessarily have to be equal in length to AC. We can draw many perpendicular lines extending inwards from AB and AC that will intercept BC at the same point (and one of those points will always be F). If we do shift A around, however, AF is no longer the height - it will become a hypotenuse of the height segment perpendicular to BC.
Ah, I did forget to disprove that it's a rectangle, but the double lines on BF and CF mean they are equal, so F is in the midpoint of BC, so AEFD has to be a square. (If that weren't the case, then there would be no way to verify or disprove the question.)
... It does make it a lot longer than the "short version" I tried to offer up, though....
If AFC were equilateral, then you couldn't fit a ∠AEF that was a right triangle in. (And, to use Asashio's logic, why would Kashima go out of her way to declare those points and make the dotted lines with marks for right angles if they weren't supposed to be used to find the answer?)
Since we already know that three of the angles of AEFD are 90 degrees, then AEFD is a square, because the sum of the angles have to add up to 360. So long as AB, AC, and BC are all straight lines with D, E, and F being points along those lines, then we can be absolutely certain of the exact angles, with ∠AFB being a right angle, and ∠ABF and ∠BAF being 45 degree angles.
The right triangles are formed by the bisection of ∠AFC, splitting the 60-degree angle into a pair of 30-degree angles. ∠AFE is 30 degrees, ∠FAE is 60, and ∠AEF is the right angle. The right angles can be used to establish parallel lines, and the rectangle itself. The (Euclidean) quadrilateral with the property of 4 right angles is the rectangle. A square is a particular type of rectangle with sides of equal length.
Edit:
FJH said: I think the trick here is application of the altitude rule of triangle theorems. The height of a right triangle is a mean proportional between the segments of the hypotenuse that it divides. In the case of a 45-90-45, that could be used as proof for AF==BF==CF, since AF would be the altitude of BAC in that case, perpendicular to BC, and BC is perfectly divided in half by it. That works in the ideal circumstance but can we extend it for a right triangle where A isn't directly above the midpoint of BC?
An altitude has to be perpendicular to the line segment it intersects, but not necessarily intersect at the midpoint. If we permit A to be rotated around F, AF would not necessarily intersect BC at a right angle.
Thinking about this a bit further it's really simple. AF and BF are equal in length (CF too, but let's put that aside for now). It means AF is exactly at midpoint of BC, now we only need to prove it's perpendicular. The easier way without measuring angles would be measuring sides with altitudes, segments AD equaling EF and AE equaling DF. That should give enough evidence AF is perpendicular to angle BFC and everything is 90 or 45 degrees angled.
After searching, I was able to find a pretty short proof using vectors, which is overkill, but also really easy for me to understand. The simpler geometric proof I also had to search for, and it took a lot more thinking for me to wrap my head around.
In intuitive terms, you can construct a right triangle out of four similar ones with half the length for each side (one way of doing so looks like a triforce, the other looks like the one in the picture). An angle identity for parallel lines shows that a smaller triangle sharing an angle with the big triangle is similar, and SAS similarity proves that the triangles containing the bisector are congruent to those in the corners.
A few tricky bits: AF does not bisect angle BAC generally ADFE is not a square generally Any right triangle works, no 45 degree angles needed
Yes!In the first place, are problems like this even needed to fight the enemy, hmmm?...please prove it more mathematically.You're quick off the mark, Asashio-san!
Go ahead.Using any method that you can think of,If you don't know the answer, then please don't raise your hand from the start.please prove how it is that AF=BF=CF in this situation.I believe the fact that this question is a problem to be solved indicates that, paradoxically, it must be true and can be solved.
Thus, it can be proven that AF=BF=CF.