<A is right, AB=AC, D , E in BC such that BD:DE:EC = 3:5:4 prove <DAE=45^o

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<A is right, AB=AC, D , E in BC such that BD:DE:EC = 3:5:4 prove <DAE=45^o

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#1 h Sep 30, 2018, 10:33 AM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle with $\angle A=90^{\circ}$ and $AB=AC$ . Let $D$ and $E$ be points on the segment $BC$ such that $BD:DE:EC = 3:5:4$ . Prove that $\angle DAE= 45^{\circ}$

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arqady

#2 h Sep 30, 2018, 11:23 AM • 2 Y

Y by Adventure10, Mango247

parmenides51 wrote:

Let $ABC$ be a triangle with $\angle A=90^{\circ}$ and $AB=AC$ . Let $D$ and $E$ be points on the segment $BC$ such that $BD:DE:EC = 3:5:4$ . Prove that $\angle DAE= 45^{\circ}$

Let $\Delta BFA\cong CEA$ such that $F$ and $C$ be placed in the different sides respect to $AB$ .
Thus, $\measuredangle FBD=45^{\circ}+45^{\circ}=90^{\circ},$ which says $FD=DE$ , $\Delta FAD\cong\Delta EAD$ .
Id est, $\measuredangle DAE=\frac{1}{2}\measuredangle FAE=\frac{1}{2}\measuredangle BAC=45^{\circ}$ and we are done!

This post has been edited 3 times. Last edited by arqady, Sep 30, 2018, 11:26 AM

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PROF65

#3 h Sep 30, 2018, 4:23 PM • 2 Y

Y by Adventure10, Mango247

Let $K,M$ be the foot of $D$ on $AB$ and the midpoint of $BC$ we have $\frac{KD}{KA}=\frac{BK}{KA}=\frac{BD}{DC}=\frac13,\frac{ME}{MA}=\frac13$ thus $AKD \sim AME$ thus $\angle KAD= \angle MAE$ therefore the result follows .
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This post has been edited 1 time. Last edited by PROF65, Sep 30, 2018, 4:24 PM

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#4 h Oct 3, 2018, 4:28 PM • 2 Y

Y by Adventure10, Mango247

Let $F$ reflection of $B$ in $AD$ ; $A$ is the circumcenter of $\triangle BFC$ , and subsequently $\angle BCF+\angle CBF=45^\circ$ . By symmetry $DF=BD,\angle CDF=2\angle DBF$ . If we draw the perpendicular bisector of $CF$ , it passes through $A$ and it cuts $BC$ at $E'$ , so that $CE'=FE'$ and $\angle DEF=2\angle DCF$ , that is, $\angle FED+\angle EDF=90^\circ$ . We then have $DF^2+FE'^2=DE'^2$ ; with $DE'+FE'=9$ , we get $DE'=5, FE'=4$ , that is, $E'\equiv E$ and indeed, $\angle DAE=45^\circ$ , as constructed.

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#5 h Oct 30, 2018, 2:58 PM • 1 Y

Y by Adventure10

Construction: Construct a point $F$ on $BC$ such $\boxed {EC=BF}$ and $M$ on $BC$ such that $\boxed {BM=MC}$ and $AM$ perpendicular to $BC$

$AM=6x ; EM=2x ; AE=2\sqrt {10} x \text { and } DM=3x ; AM=6x ; AD=3\sqrt {5} x$
Hence, $\sin \angle DAE = \sin (\angle MAD +\angle MAE) = \sin \angle MAD \cdot \cos \angle MAE + \cos \angle MAD \cdot \sin \angle MAE = \frac{1}{\sqrt {10}} \cdot \frac {2}{\sqrt {5}} + \frac {1}{\sqrt {5}} \cdot \frac {3}{\sqrt {10}} =\frac {1}{\sqrt {2}} \implies \angle DAE =45^{\circ}$ Hence Proved

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