Bosnia and Herzegovina TST 1996 Day 1 Problem 3

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Bosnia and Herzegovina TST 1996 Day 1 Problem 3

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Source: Bosnia and Herzegovina Team Selection Test 1996

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#1 h Sep 20, 2018, 11:31 AM • 2 Y

Y by Adventure10, Mango247

Let $M$ be a point inside quadrilateral $ABCD$ such that $ABMD$ is parallelogram. If $\angle CBM = \angle CDM$ prove that $\angle ACD = \angle BCM$

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#2 h Oct 30, 2019, 5:01 PM • 1 Y

Y by Adventure10

Nice problem, although too easy for a G3 nowadays

Let $E$ and $F$ be the reflections of $M$ over the midpoints of $BC$ and $CD$ respectively. Note that $BMCE$ and $DMCF$ are parallelograms.
We wish to prove $\angle BCM = \angle ACD$ and have $\angle CBM = \angle CDM$ . It is easy to see $DF=CM=BE$ and $AD=BM=CE$ . From $AF=BC$ we get $\triangle ADF \cong BEC$ . $\angle BCM = \angle EBC$ and $\angle CDM= \angle DCF = \angle MBC = \angle BCE = \angle DAF$ this means $DACF$ is cyclic which means $\angle ACD=\angle DFA = \angle BCM$ .

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#3 h Oct 31, 2019, 8:30 AM • 1 Y

Y by Adventure10

$\left. \begin{array}{l} DE// = AC\mathop {}\limits_{} \\ CE// = BM\mathop {}\limits_{} \\ CT// = BM \\ \end{array} \right\}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\limits_{}^{} C\hat ME = M\hat DC\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\limits_{}^{} M\hat DE = M\hat CD\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\limits_{}^{} B\hat CM = A\hat CD$

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This post has been edited 1 time. Last edited by Tsikaloudakis, Oct 31, 2019, 8:32 AM

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#4 h Oct 31, 2019, 5:02 PM • 1 Y

Y by Adventure10

another solution for ;
first: use trigonometry and solve this second: extend line $CM$ and $BM$ until intersect line $AB and AC$ at $M,N$ respectively then show that quadrilateral $AKML$ is cyclic and then according power of point c respect the circumcircle of $AKML$ COMPUTE $CL$ and then show that
$BMC$ and $DLC$ similar it means $<DCP=<MCB$ as desired.

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#5 h Oct 31, 2019, 5:32 PM • 1 Y

Y by Adventure10

Draw lines through $C$ parallel to $AB$ and $AD$ (they are lines through $AB \cap DM$ and $BM \cap AD$ ). By condition they are isogonal wrt $\angle BCD$ . Result immediately follows from isogonal lemma.

This post has been edited 1 time. Last edited by zuss77, Oct 31, 2019, 5:33 PM

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