angle chasing inside a parallelogram, <ABC=105, <CMB=105, BMC equilateral

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parmenides51

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parmenides51

#1 h Jul 14, 2020, 7:09 AM

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In the parallelogram $ABCD$ , $\angle ABC = 105^o$ . It is known that inside this parallelogram there is such a point $M$ that the triangle $BMC$ is equilateral and $\angle CMD = 135^o$ . Let the point $K$ be the midpoint of the side $AB$ . Find the measure of the angle $BKC$ .

(Vyacheslav Yasinsky)

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vanstraelen

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vanstraelen

#2 h Jul 14, 2020, 11:01 AM

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Let $A(0,0),B(b,0)$ ; we use $\tan 75^{\circ}=2+\sqrt{3}$ .

$D(d,(2+\sqrt{3})d)$ and $C(b+d,(2+\sqrt{3})d)$ .
$K(\frac{k}{2},0)$ and $M(\frac{b}{2},\frac{b}{2})$ .

$\angle DCB=75^{\circ}, \angle MCB=60^{\circ}$ , then $\angle DCM=15^{\circ}$ .
Given $\angle CMD=135^{\circ}$ , so $\angle CDM =30^{\circ} \Rightarrow d=\frac{(\sqrt{3}-1)b}{4}$ .

Calculating the slope of the line $KC\ :\ m_{KC}=1 \Rightarrow \angle CKB=45^{\circ}$ .

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sunken rock

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sunken rock

#3 h Jul 14, 2020, 3:54 PM

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parmenides51 wrote:

In the parallelogram $ABCD$ , $\angle ABC = 105^o$ . It is known that inside this parallelogram there is such a point $M$ that the triangle $BMC$ is equilateral and $\angle CMD = 135^o$ . Let the point $K$ be the midpoint of the side $AB$ . Find the measure of the angle $BKC$ .

Some info missing, if $M\equiv D$ , the required angle is some $40.9^\circ$

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parmenides51

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parmenides51

#4 h Jul 14, 2020, 5:44 PM • 1 Y

Y by Mango247

parmenides51 wrote:

In the parallelogram $ABCD$ , $\angle ABC = 105^o$ . It is known that inside this parallelogram there is such a point $M$ that the triangle $BMC$ is equilateral and $\angle CMD = 135^o$ . Let the point $K$ be the midpoint of the side $AB$ . Find the measure of the angle $BKC$ .

the original wording in Ukrainian is

Quote:

У паралелограмі АВСD, <АВС = 105°. Відомо, що всередині цього паралелограма існує така точка М, що трикутник ВМС рівносторонній та <СМБ = 135^o Нехай точка К - середина сторони АВ. Знайдіть градусну міру кута ВКС.

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sunken rock

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sunken rock

#5 h Jul 14, 2020, 5:53 PM

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sunken rock wrote:

parmenides51 wrote:

In the parallelogram $ABCD$ , $\angle ABC = 105^o$ . It is known that inside this parallelogram there is such a point $M$ that the triangle $BMC$ is equilateral and $\angle CMD = 135^o$ . Let the point $K$ be the midpoint of the side $AB$ . Find the measure of the angle $BKC$ .

Some info missing, if $M\equiv D$ , the required angle is some $40.9^\circ$

Sorry, I did not read $\angle ABC=105^\circ$ !!

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sunken rock

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sunken rock

#6 h Jul 14, 2020, 6:10 PM

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If $O$ was the circumcenter of $\triangle CDM$ , then $\triangle COM$ is equilateral, $OD=AD, \angle ODM=\angle ADM$ , thus $\triangle AMD\cong\triangle OMD$ (s.a.s.), hence $AM=AD,\angle AMB=90^\circ\implies MK\bot AB, MK=BK$ , therefore $BCMK$ is a kite and we are done, $\angle BKC=45^\circ$ .

Best regards,
sunken rock

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Ucchash

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Ucchash

#7 h Jul 15, 2020, 10:54 AM • 2 Y

Y by Mango247, Mango247

Draw a triangle ODC such that OM||AD and OM||BC.
so,ODAM and MBCO are parralleogram
OB and DC intersects at L
DOLM cyclic.
OB and MC intersects at T.
<BTC=<BTM=90
again, In triangle MLC
MT=TC so
<MLT=<TLC=75
through angle chase
<MAB=<MBA=45
so,MK is the perpendicular on AB.
so,<MKB=90
<KMB=<MBK=45
But MC=BC
and MK=KB
BCMK is a kite
so,<BKC=45.

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Ucchash

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Ucchash

#8 h Jul 15, 2020, 11:15 AM • 1 Y

Y by Mango247

Draw a triangle ODC such that OM||AD and OM||BC.
So,OCBM is a parrellogram
OB and Mc intersect at T
and<TBC=<TBM=30
Let <BKC=x
<BCK=75-x and <MCK=x-15
CK, BT intersects at U
draw MU.
<UMC=<UCM=x-15
In triangle MCU and BCU
CU/MU=CU/BU
or,sin(x-15)/sin(210-2x)=sin30/sin(75-x)
x=45=<BKC

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