similar triangles wanted, trapezoid, right angles (2012 NZOMC Camp Sel. S2)

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similar triangles wanted, trapezoid, right angles (2012 NZOMC Camp Sel. S2)

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#1 h Jan 4, 2021, 7:22 AM

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Let $ABCD$ be a trapezoid, with $AB \parallel CD$ (the vertices are listed in cyclic order). The diagonals of this trapezoid are perpendicular to one another and intersect at $O$ . The base angles $\angle DAB$ and $\angle CBA$ are both acute. A point $M$ on the line sgement $OA$ is such that $\angle BMD = 90^o$ , and a point $N$ on the line segment $OB$ is such that $\angle ANC = 90^o$ . Prove that triangles $OMN$ and $OBA$ are similar.

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#2 h Jan 4, 2021, 7:19 PM

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Given $A(0,0),B(b,0),C(c,a),D(d,a)$ .
The condition $AC \bot BD$ gives $d=b-\frac{a^{2}}{c}$ .
Point $O(\frac{bc^{2}}{a^{2}+c^{2}},\frac{abc}{a^{2}+c^{2}})$ .
The line $AC\ :\ y=\frac{a}{c}x$ with $M(m,\frac{am}{c})$ .
The condition $BM \bot DM$ gives $m=\frac{bc^{2}-\sqrt{a^{2}bc(a^{2}-bc+c^{2})}}{a^{2}+c^{2}}$ .
The line $BD\ :\ y=-\frac{c}{a}(x-b)$ with $N(n,\frac{c(b-n)}{a})$ .
The condition $AN \bot CN$ gives $n=\frac{bc^{2}+\sqrt{a^{2}bc(a^{2}-bc+c^{2})}}{a^{2}+c^{2}}$ .
I thought $MN\ //\ AB$ , but $y_{M}\ \neq\ y_{N}$ .
After calculations $\frac{ON}{OA}=\frac{OM}{OB}=\frac{MN}{AB}$ , so $\triangle OMN \sim \triangle OAB$ .

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