equal angles lead to equal segments, projections related

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equal angles lead to equal segments, projections related

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Source: Australian MO 1983

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#1 h Jul 9, 2019, 7:36 PM • 2 Y

Y by Adventure10, Mango247

$ABC$ is a triangle and $P$ is a point inside it such that $\angle PAC=\angle PBC$ . The perpendiculars from $P$ to $BC$ and $CA$ meet these sides at $L$ and $N$ respectively, and $D$ is the midpoint of $AB$ . Prove that $DL = DN$ .

This post has been edited 1 time. Last edited by parmenides51, Jul 10, 2019, 3:08 PM
Reason: typo see #2

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#2 h Jul 10, 2019, 2:52 PM • 1 Y

Y by Adventure10

parmenides51 wrote:

$ABC$ is a triangle and $P$ is a point inside it such that $\angle PAC=\angle PBC$ . The perpendiculars from $P$ to $BC$ and $CA$ meet these sides at $L$ and $N$ respectively, and $D$ is the midpoint of $AB$ . Prove that $DL = DM$ .

Did you mean $DL=DN$ .

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#3 h Jul 10, 2019, 3:09 PM • 2 Y

Y by Adventure10, Mango247

yes, corrected thanks, there was a typo also on my source, the Australian Mathematical Olympiads 1979 - 1995 book

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#4 h Jul 10, 2019, 8:17 PM • 3 Y

Y by zuss77, Adventure10, Mango247

Let $U$ , $V$ be the midpoints of $AP$ , $BP$ , respectively. We will show that the triangles $DUN$ and $DVL$ are congruent.
By Thales, we have $UN=UP$ , and since $DVPU$ is a parallelogramm, also $UP=DV$ . Thus $UN=DV$ , and similarly $VL=DU$ . Moreover, from Thales we get $\angle PUN = 2\angle PAN = 2\angle LBP = \angle LVP$ . But since $DVPU$ is a parallelogramm, we also have $\angle PVD = \angle DUP$ . Consequently $\angle DUN = \angle LVD$ , and we are done.

This post has been edited 1 time. Last edited by Gryphos, Jul 12, 2019, 7:35 AM
Reason: N and L were swapped.

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#5 h Jul 11, 2019, 10:00 AM • 2 Y

Y by Adventure10, Mango247

Can someone provide solution with moving points? I'm not comfortable in this field. Can it be something like this? : If angles in question change and $P$ moves on hyperbola, then we need to check four cases. And in cases when $P=A,B,C$ and orhocenter the statement holds true.

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#6 h Jul 11, 2019, 2:27 PM • 1 Y

Y by Adventure10

Gryphos wrote:

Let $U$ , $V$ be the midpoints of $AP$ , $BP$ , respectively. We will show that the triangles $DUL$ and $DVN$ are congruent.
By Thales, we have $UL=UP$ , and since $DVPU$ is a parallelogramm, also $UP=DV$ . Thus $UL=DV$ , and similarly $VN=DU$ . Moreover, from Thales we get $\angle PUL = 2\angle PAL = 2\angle NBP = \angle NVP$ . But since $DVPU$ is a parallelogramm, we also have $\angle PVD = \angle DUP$ . Consequently $\angle DUL = \angle NVD$ , and we are done.

$N$ and $L$ are inverted in your proof.

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#7 h Jul 12, 2019, 7:16 AM • 1 Y

Y by Adventure10

Let $K$ be foot of perpendicular from $P$ on $AB$ .
$ANPK$ and $BLPK$ are cyclic $\implies \angle AKN = \angle BKL$ . (*)
Due to equal angles in condition, isogonal conjugate of $P$ lays on perpendicular bisector of $AB$ and it's well-known that pedal triangles of isogonal conjugates have the same circimcircle. In other words $KDLN$ - cyclic. Now, with (*), chords $DL$ and $DN$ are equal.

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