AE = AC, if <ACP =<ABC, line reflection, circumcircle related

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AE = AC, if <ACP =<ABC, line reflection, circumcircle related

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Source: 2018 Romania JBMO TST 1.3

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#1 h Jun 1, 2020, 9:13 PM

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Let $ABC$ be a triangle with $AB > AC$ . Point $P \in (AB)$ is such that $\angle ACP = \angle ABC$ . Let $D$ be the reflection of $P$ into the line $AC$ and let $E$ be the point in which the circumcircle of $BCD$ meets again the line $AC$ . Prove that $AE = AC$ .

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#2 h Jun 1, 2020, 11:50 PM

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$\angle ACD= \angle ACP = \angle ABC$ and $\angle CAD= \angle CAB$ so $\triangle ABC \sim \triangle ACD \to AC^2=AD*AB$
$\angle EBD=\angle ACD=\angle ABC$ so $\angle EBA= \angle DBC= \angle DEC$ $\angle EAD= \angle EAB$ so $\triangle EAB \sim \triangle DAE \to AE^2=AB*AD$
So $AE=AC$

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Mahdi_Mashayekhi

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#3 h Jan 5, 2022, 6:54 AM

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Let AB meet circle at S.
∠ACD = ∠ACP = ∠ABC = ∠AES so SE = DC = PC.
∠CPA = ∠ACB = ∠ASE and ∠ACP = ∠ABC = ∠AES and SE = PC so triangles AES and APC are congruent so AE = AC.
we're Done.

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#4 h Dec 23, 2022, 9:19 PM • 1 Y

Y by erkosfobiladol

$\angle ACP=\angle ABC$ So $AC^2=AP.AB$ We will show $AE^2=AP.AB$ This means we must show that $\angle ABE=\angle PEA$
If $\angle ACP=\alpha$ $\angle APD=\beta$ $\angle ADE=\theta$ , we can say that $\angle PDC=90-\alpha$ so $\angle EDC=90+\beta+\theta-\alpha$ . $EBCD$ is cyclic so $\angle EBC=90+\alpha+\beta-\theta$ We know that $\angle ABC=\alpha$ $\implies$ $\angle EBA=90-\beta-\theta$
$\beta=\angle ADP=\angle DPA$ and $\theta=\angle ADE=\angle APE$ $\implies$ $\angle PEA=90-\beta-\theta$

This post has been edited 4 times. Last edited by bin_sherlo, Dec 24, 2022, 7:21 PM

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