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Consider parallelogram $ABCD$ ($AB\parallel CD$, $BC\parallel AD$) as shown:

Parallelogram ABCD

Pick a pair of its opposite vertices, say $B$ and $D$:

Opposite vertices B and D

From these vertices, construct altitudes to the adjacent sides of this parallelogram:

  • From vertex $B$ construct an altitude $BH_1$ to side $AD$.
  • From vertex $D$ construct altitude $DH_2$ to side $AB$.

Two lines containing these altitudes will always intersect as they are perpendicular to two intersecting sides of parallelogram.

Two intersecting altitudes

Let $H$ be such intersection point.

Intersection point of two altitudes

Let $\angle DHH_1$ be the smaller angle between $BH_1$ and $DH_2$.

Angle DHH1

How to express such angle in terms of interior angles of parallelogram $ABCD$?

It is evident from the drawing that this angle is equal to the acute angle of this parallelogram, but I would like a rigorous proof of that. I wonder if it always works.

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    $\begingroup$ $\angle HDH_1 = 90^\circ -\angle A$, $\angle DHH_1 = 90^\circ -\angle HDH_1$. $\endgroup$ Commented Aug 10, 2024 at 8:52
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    $\begingroup$ $\angle DHH_1+\angle H_1HH_2=180^\circ$, and in quadrilateral $AH_1HH_2$ we have $\angle A+\angle AH_1H+\angle H_1HH_2+\angle AH_2H=360^\circ$, hence $\angle A+\angle H_1HH_2=180^\circ$. Therefore, $\angle DHH_1=\angle A$. $\endgroup$ Commented Aug 10, 2024 at 8:53

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The comment/answers from peterwhy and Feng are immediate paths, almost like visual reflexes:

  • $\triangle{ABH_1}$ divides into similar triangles
  • $AH_1HH_2$ is cyclic

Consider too that a parallelogram contains only two angles:

  • $90-\alpha$
  • $90+\alpha$

A perpendicular meeting those angles can only form:

  • complement of (acute) $90-\alpha \rightarrow \alpha$
  • removal of $90$ from (obtuse) $90+\alpha \rightarrow \alpha$

So the angle divided out by the altitude at vertex $D$ must be $\alpha$, and the angle marked in the diagram must be $90-\alpha$, which is the original angle at $A$.

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