Let ABC be a triangle with AB=AC. The bisector of angle ACB meet AB at M. Suppose AM+MC=BC. Show that angle BAC = 100 degrees? Thank you!

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Let ABC be a triangle with AB=AC. The bisector of angle ACB meet AB at M. Suppose AM+MC=BC. Show that angle BAC = 100 degrees? Thank you!

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Answer 1 · 2018-01-14T05:26:49

see explanation.

Explanation:

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Let $∠ A B C and ∠ A C B be 2 β, \Rightarrow ∠ A C M = ∠ M C B = β$ $angleABC and angleACB " be " 2beta, => angleACM=angleMCB=beta$
Mark point $D$ $D$ on $B C$ $BC$ , such that $C D = C M$ $CD=CM$ .
Draw a line $M E$ $ME$ , parallel to $B C$ $BC$ , as shown in the figure,
$\Rightarrow ∠ E M C = ∠ M C B = β$ $=> angleEMC=angleMCB=beta$ ,
and $∠ A M E = ∠ A E M = 2 β$ $angleAME=angleAEM=2beta$
let $C M = x, A M = y and M B = z$ $CM=x, AM=y and MB=z$ ,
$\Rightarrow C D = x, A E = y and E C = z$ $=> CD=x, AE=y and EC=z$ ,
In $DeltaEMC$ , as $angleEMC=angleECM=beta, => DeltaEMC$ is isosceles,
$=> EM=EC=z$
As $BC=AM+MC=BD+CD, => y+x=BD+x, => BD=y=AM$
Now, as $BM=ME=z, BD=MA=y$ , and included angle $angleDBM=angleAME=2beta$ ,
$=> DeltaDBM and DeltaAME$ are congruent,
$=> angleDMB=angleAEM=2beta$
Now, $angleCDM=angleDBM+angleDMB=2beta+2beta=4beta$
As $CD=CM, => angleCMD=angleCDM=4beta$ ,
In $DeltaCDM, 4beta+4beta+beta=9beta=180^@$ ,
$=> beta=20^@$
Hence, $angleBAC=180-2beta-2beta=180-4beta=180-(4xx20)=100^@$

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Let ABC be a triangle with AB=AC. The bisector of angle ACB meet AB at M. Suppose AM+MC=BC. Show that angle BAC = 100 degrees? Thank you!

1 Answer

Explanation:

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