Please help mee!!!!!! In triangle $ABC$, $BAC=90^\circ$, $M$ is its middle $ [BC]$ and $BD\perp AM$, $D\in(AC)$. Prove the equivalence: :: $$ ACB = 30^\circ \iff BD = 2\cdot MD $$
You can see my drawing down and also my ideas...
As you can see I could demonstrate the fact that if $ACB$ is $30^\circ$ then $BD=2\cdot MD$ But I don't know how to do the opposite of it. I know that usually the idea is the same...but no idea came into my mind. Any idea is welcome! Thank you!!!!
On
I am using the same figure used by sirous since I have no means to draw a diagram now:
Due to the mid-point theorem, we have that $EM||AC$. Also, we have $AE = BD/2 = MD$. Thus, quad. $AEMD$, is a parallelogram or an isosceles trapezium. Suppose it is an isosceles trapezium. Then $\angle AEM = \angle EMD$.
But $\angle AEM = \angle DEA + \angle DEM = \angle DEA + \angle EMD > \angle EMD$. We arrive at a contradiction, and hence $AEDM$ is a parallelogram.
Since the diagonals are perpendicular, it is also a rhombus. We can complete the required proof from here in many ways.
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As cab be seen in figure,E is mid point of BD, so :
$EM||AC$
$AM\bot ED$, so quadrilateral AEMD is rhombus and :
$ME=AD=ED=AE=MD=\frac {BD}2\Rightarrow BD=2 MD$
Also:
$\widehat{DBM}=30^o$
Triangle BAD and BMD are congruent, because:
$\angle ADB=\angle BDM=60^o$ , so $\angle ABD=\angle DBC$, which gives:
$\angle ABC=60^o \Rightarrow \angle ACB=30^o$
Update: Answer to your questions in your comment:
1- Itriangle BDC, the midpoints of sides BD and BC are conneted so $EM||DC$. We have:
$AE=\frac{BD}2=EM=ND=ED$
This together with facts $ED\bot MA$ and $EM||AD$ are enough to conclude that AEMD is a rhombus. So $MD||AE$.
2- If $BD=2 MD$ that means in right angle triangle BDM:
$\angle DBM=30^o \Rightarrow \angle BDM=60^o$
In rhombus AEMD. diagonal ED bisects angle ADM, so $\angle ADM=120^o$
This angle is exterior to to right angle triangle MDC, where:
$\angle DMC=90^o \Rightarrow \angle MCD=120-90=30^o $