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April 19th, 2018, 11:32 PM   #1
lua
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Proof that two line segments are equal

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Incircle of the triangle ABC touches edges BC and AC in points P and Q, respectively. Straight line that contains the center of the edge AB and which is parallel to PQ, intersects straight lines BC and AC in points D and E, respectively. Prove that AE=BD.
So far, I've done this (see picture).
From OP=OQ comes out that the triangle OPQ is isosceles triangle. This means that $\displaystyle \measuredangle OPQ = \measuredangle OQP$.
Now, $\displaystyle \measuredangle AQP = \measuredangle AQO + \measuredangle OQP = 90$/degree$ + \measuredangle OQP$ and $\displaystyle \measuredangle DPQ = \measuredangle DPO + \measuredangle OPQ = 90$\degree$ + \measuredangle OPQ$.
So, $\displaystyle \measuredangle AQP = \measuredangle DPQ$.
This means that quadrilateral EDPQ is isosceles trapezoid. This means that $\displaystyle EQ = DP$.
That's it. I can't go further.

P.S. Anybody have the idea how to write degrees sign in Latex? I copied this from online Latex equation editor where it is OK.
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Last edited by lua; April 19th, 2018 at 11:51 PM.
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April 20th, 2018, 05:29 AM   #2
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90^{\circ}=$90^{\circ}$
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April 21st, 2018, 10:20 PM   #3
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Can anybody help?
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April 24th, 2018, 08:21 AM   #4
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Draw $\displaystyle AX \parallel BC$

From the parallel lines we have $\displaystyle \angle AXE = \angle CDG = \angle CPQ $ and $\displaystyle \angle AEX = \angle CQP$.

But $\displaystyle \angle CQP = \angle CPQ$ so $\displaystyle \angle AEX = \angle AXE$ and we have that $\displaystyle AE = AX$.

Now observe that $\displaystyle \triangle AXG \cong \triangle BDG$ so we have that $\displaystyle AX = BD$.

Thus $\displaystyle AE = BD$
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April 24th, 2018, 12:19 PM   #5
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April 25th, 2018, 07:01 AM   #6
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Because I am so picky, I feel the need to point out that the two line segments are NOT "equal" as implied in the title, They are "congruent" because they have the same length: AE= BD, as stated in the problem, where AE and BD are the lengths of line segments $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$.

In mathematics to say that $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$ are equal would mean that $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$ are different names for the same line segment.
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April 25th, 2018, 09:14 AM   #7
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Originally Posted by Country Boy View Post
Because I am so picky, I feel the need to point out that the two line segments are NOT "equal" as implied in the title
Euclid would disagree!
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April 25th, 2018, 11:44 AM   #8
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Quote:
Originally Posted by Country Boy View Post
Because I am so picky, I feel the need to point out that the two line segments are NOT "equal" as implied in the title, They are "congruent" because they have the same length: AE= BD, as stated in the problem, where AE and BD are the lengths of line segments $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$.

In mathematics to say that $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$ are equal would mean that $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$ are different names for the same line segment.
How come you aren't as picky about angles? As sets of points, angles can only be congruent not equal. Their measures can be equal.
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May 5th, 2018, 11:05 AM   #9
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I am! I just thought that bitching about the line segments was enough!
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