Assume the figure ABCD is a square. Divide the segment AB in half and through the point of division (E) create a line perpendicular to AB. This line will intersect the opposite side of the square (DC) at F. Such that DF = FC.
From the point C, create a segment CG equal to CB. Connect points A and G with a line and divide the segment AG in half by a point H. Then, from the point H create a line HK which is perpendicular to the segment AG.
Because the segments AB and AG are not parallel, the lines EF and HK are also not parallel. Therefore, they will intersect at some point (call it K). Let’s connect the point K with points D, A, G and C.
Triangles KAH and KGH are congruent (equal?), because they have a common side HK, because AH = HG, and because the angles at H are right angles (both equal to 90°). [Here we are invoking a well-known theorem that says that if two triangles’ sides and the angle between those two sides are equal, the triangles are congruent.] Therefore, KA = KG.
Triangles KDF and KCF are also congruent, because they have a common side FK, because DF = FC, and because the angles at the point F are right (both equal to 90°). Therefore, KD = KC, and angle KDC is equal to angle KCD.
Besides that, DA = CB = CG.
Therefore, the sides of the triangles KDA and KCG are equal. Therefore, angles KDA and KCG are equal. Let’s subtract from them the equal angles KDC and KCD. It is obvious that the results of the subtraction should also be equal; i.e., that angle GCD = angle ADC.
But, the angle GCD is an obtuse angle (>90°). While the angle ADC is a right angle (90°).
Therefore, sometimes, obtuse angle is equal to the right angle, which is what was required to prove and what is so far from the truth.
9 comments:
hmmm
And?..
I was gchatting at the same time, but I don't want to spend more time thinking about this, it had me getting very confused. I'm not sure the source of the problem, but one place you see doesn't fit at all is when you say that KA equals KG, and KAH equals KGH, because KA should equal KB, if K is on the line of EF.
anyway, I think the problem is how you call the point of intersection K, and start drawing equal lines to D and C, and equal angles
If I had some paper and a pencil, it'd be easier.
please share the answer
KA = KG, because two triangles (KAH and KGH) are congruent, and as a result, their sides are equal. It doesn't say KA = KB. Read that part of the proof more carefully.
There is nothing special about point K. You draw a line perpendicular to AG from point H. That line must intersect line EF at some point (since AB and AG are not parallel). Call that point K and draw lines from it to the said points.
We don't postulate a priori that the lines drawn from K are equal. We prove it from congruency of the triangles. The only things we postulate a priori to be equal are the segments which we made equal by dividing a segment in half. (Plus, there is a common side; and because lines were drawn perpendicular, their formed right angles, which are obviously equal.)
I don't want to share the answer just yet, since other people may be interested in figuring it out. And the proof may be more interesting to think about that to know the answer.
As a hint, what's wrong with the proof is not the logic of the proof itself. I.e., on the first round, you should be able to go through the proof, and it should make sense.
ok, one more go at it and then I quit.
The mistake is that the intersection of EF and HK will be before the line DC, not after. I followed through with all the proofs, which are fine, but when you get to the end, you still have not proven how DCG is equal to ADC. Perhaps KCG, I don't know.
and that's the last time I'm trying
one last comment: I'm pretty sure if I were to construct the shapes and angles myself, based on the problem, I would have a different picture than the one you have
The mistake is that the intersection of EF and HK will be before the line DC, not after.
Wrong, but warm.
when you get to the end, you still have not proven how DCG is equal to ADC. Perhaps KCG, I don't know.
Triangles ADK and GCK are congruent. Therefore, their corresponding angles are equal. Therefore, angle ADK = angle GCK. If you subtract from the two angles equal angles KDC and KCD, the results should be equal. (If x = y, and k = m, then x − k = y − m) But the results are a right angle (ADC) and an obtuse angle (DCG).
one last comment: I'm pretty sure if I were to construct the shapes and angles myself, based on the problem, I would have a different picture than the one you have
Different how? Even if the exact distances are not correct on the picture, surely the logic still works, no? :)
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