Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $D$ be the midpoint of side $BC$ and $BE,CZ$ be the altitudes of the triangle $ABC$. Line $ZE$ intersects line $BC$ at point $O$. (i) Find all the angles of the triangle $ZDE$ in terms of angle $\angle A$ of the triangle $ABC$ (ii) Find the angle $\angle BOZ$ in terms of angles $\angle B$ and $\angle C$ of the triangle $ABC$
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Tags: geometry, angles, altitudes
dchen04
24.02.2020 02:33
a) We note that because $\angle BEC=\angle BZC$, points $B,C,E,Z$ are concyclic with the midpoint of $\overline{BC}$, $D$, being the center of the circle. We deduce that triangles $BDZ$, $ZDE$ and $EDC$ are isosceles. Angle chasing gives us that $\angle EZD=180^{\circ}-\angle BDZ-\angle CDE=180^{\circ}-(180^{\circ}-2\angle B)-(180^{\circ}-2\angle C)=180^{\circ}-2\angle A$ and $\angle EZD=\angle ZED=\angle A$.
b) Again, angle chasing gives us $\angle BOZ=\angle COE = 180^{\circ}-\angle CEO-\angle C=180^{\circ}-\angle ZED-\angle CED-\angle C=180^{\circ}-\angle A-2\angle C=\angle B-\angle C$.
Math-Shinai
24.02.2020 07:13
part a is the well-known 3 tangent lemma, while part b is the well-known cyclic configuration
parmenides51
30.04.2024 09:25
Let <ZBE=x, <EBC=y, <ZCB=z
BZEC is cyclic as ,<BZC=<BEC=90^o
=> <EZC=<EBC=y and <ZEB=<ZCB=z
From median of hypoteinuses we get ZD=BC/2=ED
=> ZD =DC which gives <DZC=<ZCD=z
and BD=DE which gives <BED=<EBD=y
But from triangle Sum in AZC we get: x=90^o -<A (1)
From triangle sum in ABC we get: 2x + y+z+<A=180^o , so using (1) we get y+z=a<A
So <DZE=<ZDE=y+z=<A
<ZDE=180^o -2 <A
<OBZ=180^o -<B
<OZB =180^o- (x+y+z+y)=180^o-<B-<A=<C
<BOZ=180^o-(180^o -<B +C)=<B-<C