AB <BD wanted, two right triangles, midpoint 2004 Portugal p2

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AB <BD wanted, two right triangles, midpoint 2004 Portugal p2

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#1 h Oct 13, 2020, 1:28 PM

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Consider the triangles $[ABC]$ and $[EDC]$ , right at $A$ and $D$ , respectively. Show that, if $E$ is the midpoint of $[AC]$ , then $AB <BD$ .

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This post has been edited 2 times. Last edited by parmenides51, Oct 13, 2020, 3:31 PM
Reason: added figure

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#2 h Oct 13, 2020, 2:02 PM

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$\text{[math]}$

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#3 h Oct 13, 2020, 2:13 PM

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They both have same hypo..(forgotten spelling) . ec=ae . ec >ed. Hence bd >ab

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#4 h Oct 13, 2020, 2:56 PM

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@angle bisector, is $D$ necessarily on $BC$ ?

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#5 h Oct 13, 2020, 3:31 PM

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there is a figure missing, my fault, wait

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#6 h Oct 13, 2020, 3:43 PM

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I feel like we have to use similarity but I can't figure out how

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#7 h Oct 13, 2020, 3:44 PM

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We have $\triangle CDE \sim \triangle CAB$ ...

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#8 h Oct 13, 2020, 3:45 PM

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IAmTheHazard wrote:

We have $\triangle CDE \sim \triangle CAB$ ...

I know that, but I am not sure how to use it.

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#9 h Oct 13, 2020, 3:52 PM

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Clearly, quadrilateral $ABDE$ is cyclic. Then, we know that $AB^2+AE^2=DE^2+BD^2$ . Clearly, $AE=CE$ so $AB^2-BD^2=DE^2-CE^2$ . Clearly, $DE^2-CE^2$ is negative so the claim is proven.

naw bro
cyclic quads is all you need

This post has been edited 1 time. Last edited by Pleaseletmewin, Oct 13, 2020, 3:52 PM

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#10 h Oct 13, 2020, 8:51 PM

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Nice one @above!

You don't even need to recognize the cyclic quadrilateral. Pythag on the shared hypotenuse may be more intuitive.

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#11 h Nov 28, 2020, 3:54 PM

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Let $M$ be the midpoint of $BC$ , thus we need to prove that $BD=CM+MD>2EM$ .
Let $CM=c$ , $EM=a$ and $MD=x$ , let $\angle ACB=\alpha$ , thus
$a=c\sin{\alpha}$ and $x=a\sin{\alpha}$ , we get that
$c\sin^2{\alpha}+c>2c\sin{\alpha}\Longleftrightarrow c(\sin{\alpha}-1)^2>0.$ $\sin{\alpha}-1=0$ cannot hold, because $\alpha<90^\circ$ .

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