Kazakhstan 2012 ( Grade 10 problem 2)

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Kazakhstan 2012 ( Grade 10 problem 2)

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cyclic quadrilateral

perpendicular bisector

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#1 h May 20, 2012, 5:36 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be an inscribed quadrilateral, in which $\angle BAD<90$ . On the rays $AB$ and $AD$ are selected points $K$ and $L$ , respectively, such that $KA = KD, LA = LB$ . Let $N$ - the midpoint of $AC$ .Prove that if $\angle BNC=\angle DNC$ ,so $\angle KNL=\angle BCD$

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#2 h May 23, 2012, 3:13 AM • 1 Y

Y by Adventure10

Let $X,Y$ be on rays $AB,AD$ such that $XD\perp AD$ and $YB\perp AB$ . Then $K,L$ are midpoints of $AX,AY$ , so now it suffices to show $\angle XCY=\angle BCD$ . From $\angle BNC=\angle DNC$ , $ABCD$ is harmonic, so $\frac{BC}{CD}=\frac{BA}{AD}$ . Since $\angle YBC=\angle XDC$ and $\frac{BC}{CD}=\frac{BA}{AD}=\frac{BY}{DX}$ , we have $\triangle BCY\sim\triangle DCX$ so $\angle BCY=\angle DCX$ and the result follows.

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#3 h May 24, 2012, 6:04 PM • 2 Y

Y by Adventure10, Mango247

Let Γ be the circumcircle of quadrilateral ABCD, O be its circumcenter, alpha = ∠LAB = ∠LBA = ∠KDA, E and F be the midpoints of AB and AD, respectively. Now let beta = ∠ELA = ∠ELB = ∠FKA = ∠FKB. We see that alpha + beta = 90°. Because ∠ELB = ∠OLB = ∠FKA = ∠OKA, BKLO is cyclic. Now extend BN to meet Γ at D’. Since N is the midpoint of AC and ∠BNC = ∠DNC, D’ is the point symmetry of D with respect to ON. In other words, arc AD’ = arc CD. Therefore, ∠BAD which subtends arc BD equals ∠BNC = ∠DNC. This implies that ∠OND = beta.

From there, we get ∠ONB + ∠OLB = 2(alpha + beta) = 180°, and BLON is also cyclic. Combining with BKLO being cyclic, we conclude that BKLON is cyclic, and ∠KNL = ∠KBL (both subtend arc KL) = 180° – ∠LBA = 180° – alpha = ∠BCD.

Sorry, I am not familiar with LaTex. No time to learn.

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hyperspace.rulz

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hyperspace.rulz

#4 h Jan 20, 2013, 11:21 AM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the circumcentre of cyclic quadrilateral $ABCD$ . Let $\angle DAB=\alpha$ . We are given that $\angle BNC=\angle DNC$ . Note that:

$\angle DOB=2\alpha$
$\angle DKB=\angle DKA=180^0-\angle KAD-\angle KDA=180^0-2\alpha$
$\angle DLB=\angle BLA=180^0-\angle LAB-\angle LBA=180^0-2\alpha$

Hence $K, L$ both lie on the circumcircle of $\triangle DOB$ .

Now note that we are given that $NC$ internally bisects $\angle DNB$ . Also, since $N$ is the midpoint of $AC$ , we have that $ON$ is perpendicular to $NC$ i.e. that $ON$ is the external angle bisector of $\angle DNB$ . Also, $O$ clearly lies on the perpendicular bisector of $DB$ , since $OD=OB$ . Therefore, $O$ is the intersection of the perpendicular bisector of $BD$ with the external angle bisector of $\angle DNB$ , and hence lies on the circumcircle of $\triangle DNB$ . Hence $N$ lies on the circumcircle of $\triangle DOB$ , as do $K$ and $L$ . Therefore $L, D, O, N, B, K$ are all concyclic.

Now $\angle KNL=\angle KDL=180^0-\angle KDA=180^0-\angle KAD=\angle BCD$ , and thus we are finished.

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