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The Question Is:

BH is an altitude of triangle ABC. Let K and P be points symmetric to H with respect to AB and BC respectively. Prove that KP intersects AB and BC at the bases of the other two altitudes of triangle ABC.

The second line of the problem really throws me off. Any help in understanding this question would be greatly appreciated!

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    $\begingroup$ Say that perpendicular from H to AB intersects AB at point X. Then K is the point on that perpendicular such that |HX| = |KX|. That is, K is the mirror image of H reflected in the line AB. P is defined similarly. Does this make it clear? $\endgroup$ – saulspatz Oct 20 '19 at 20:16
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    $\begingroup$ Thank you so much! $\endgroup$ – Mathview Oct 20 '19 at 20:19
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This is how the points are situated:

enter image description here

And this is the claim you have to show:

enter image description here

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  • $\begingroup$ I think you switched the positions of K and P. $\endgroup$ – user706791 Oct 20 '19 at 20:30
  • $\begingroup$ do you know how to prove this statement? $\endgroup$ – user706791 Oct 20 '19 at 20:35
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enter image description here

Refer to the diagram for the proof of the statement:

Since CYXA is cyclic due to the altitude lines HX and AY, $ \angle BCA = 180 -\angle YXA $. BXHC is cyclic due to the altitude lines BH and CX and

$$ \angle HXA = 180 - \angle BXH = \angle BCA = 180 -\angle YXA $$

Also, because P is the mirror image of H respect to XA,

$$\angle PXA = \angle HXA = 180 - \angle YXA$$

Thus, the points P, X and Y are colinear.

Similarly, the points K, Y and X are colinear. Therefore, KP intersects AB and BC at the bases of the other two altitudes of ABC.

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