AD = DU if <CBA = 60^o, <CBU = 45^o, U circumcenter

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AD = DU if <CBA = 60^o, <CBU = 45^o, U circumcenter

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Source: 2021 Austrian Mathematical Olympiad Junior Regional Competition , Problem 2

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#1 h Sep 12, 2021, 5:19 PM

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A triangle $ABC$ with circumcenter $U$ is given, so that $\angle CBA = 60^o$ and $\angle CBU = 45^o$ apply. The straight lines $BU$ and $AC$ intersect at point $D$ . Prove that $AD = DU$ .

(Karl Czakler)

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#2 h Sep 12, 2021, 5:33 PM

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Sol

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#3 h Sep 12, 2021, 5:37 PM

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Since $\angle CBU = 45^\circ$ , we find that $\angle BUC=90^\circ$ and $\angle BAC=45^\circ$ .
$\angle UAD=30^\circ$ and $\angle ADU=180^\circ-\angle UDC=180^\circ-(90^\circ-\angle UCD)=120^\circ$ . Thus, $\angle AUD=30^\circ=\angle UAD$ .
Then, $|AD|=|DU|$ , as desired.

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#4 h Sep 12, 2021, 6:03 PM

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What they ask us is equivalent to:
$\angle AUD= \angle DAU$
Notice that $\angle ABD=15$
$\implies \angle AUD=30$
With chasing angles
$\angle DAU =30.$

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#5 h May 10, 2024, 10:11 AM

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Notation : $(XY)$ stands for small arc $XY$

Let $BU$ intersect the circumcircle at $M$
$\angle AUD=(AM)=2\angle ABU =2*15^o=30^o$
$(AB)=\angle BUA=180^o-2\angle ABU=180^o-2*15^o=180^o-30^o=150^o$
$(MC)=2\angle MBC= 2*45^o=90^o$
Due to lemma $\angle UOA= (AB)/2 + (MC)/2=(150^o+90^o)/2 =120^o$
So by angles triangle sum in $UAO,$ $\angle UAO=30^o=\angle AUD \Rightarrow AD=UD$

Lemma: If two chords $XY$ and $ZW$ intersect inside their circle at point $K$ , $\angle XKW= ( (YW)+(ZY) ) /2$

This post has been edited 3 times. Last edited by parmenides51, May 10, 2024, 7:27 PM

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#6 h Nov 11, 2024, 10:35 AM

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$\measuredangle CBU+\measuredangle UBA=\measuredangle CBA\Rightarrow \measuredangle UBA=15$
$\triangle CBU$ isosceles $\Rightarrow \measuredangle UCB=\measuredangle CBU=45$ and $\measuredangle BUC=90$
$\triangle ABU$ isosceles $\Rightarrow \measuredangle BAU=\measuredangle UBA=15$ and $\measuredangle AUB=150$
$\measuredangle AUB+\measuredangle BUC+\measuredangle CUA=360\Rightarrow \measuredangle CUA=120$
$\triangle CAU$ isoscoles $\Rightarrow \measuredangle UAC=\measuredangle ACU=30$ because $\measuredangle UAC+\measuredangle ACU+\measuredangle CUA=180$ (Also note that $\measuredangle UAC=\measuredangle UAD$ because $C;D;A$ is colinear, so $\measuredangle UAD=30$ )
$\measuredangle DUA+\measuredangle AUB=\measuredangle DUB=180\Rightarrow \measuredangle DUA=30$
$\measuredangle DUA=30=\measuredangle UAD\Rightarrow AD=DU$

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