AE=2EB inside a right isosceles triangle (Singapore Junior 2019)

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AE=2EB inside a right isosceles triangle (Singapore Junior 2019)

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#1 h Jul 10, 2019, 11:29 AM • 2 Y

Y by Adventure10, Dumb_at_math_1729

In the triangle $ABC, AC=BC, \angle C=90^o, D$ is the midpoint of $BC, E$ is the point on $AB$ such that $AD$ is perpendicular to $CE$ . Prove that $AE=2EB$ .

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#2 h Jul 10, 2019, 2:26 PM • 2 Y

Y by Adventure10, Mango247

Solution.

This post has been edited 1 time. Last edited by Bashy99, Jul 10, 2019, 2:27 PM
Reason: missing $

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#3 h Jul 10, 2019, 3:30 PM • 1 Y

Y by Adventure10

Let $AC=BC=1, BD=CD=\frac{1}{2},AD=\frac{\sqrt 5}{2}$ . $AB=\sqrt 2$
Let $F$ lies on $AB$ & $BF=\frac{\sqrt 2}{3}.O$ is the intersection of $AD$ & $CF$
Now, $\text{cos} \angle BCF$ $=\frac{CA}{AB}=\frac{2}{\sqrt 5}$
Apply stewart's theorem as $CF=d$ , as a chevian,
$AC^2\cdot BF+BC^2\cdot AF=AB(d^2+AF\cdot BF)$
$\implies d=\frac{\sqrt 5}{3}$ .
In $\triangle BCF$ apply law of cosine and get $\text {cos}\angle BCF=\frac{2}{\sqrt 5}$
$\implies \text {cos}BCF=\frac{2}{\sqrt 5}=\text{cos} \angle CAB$
So, $\triangle DAO$ is similar to $\triangle ACD$ .
So, $AD\perp CF$ . But we know $AD\perp CE$ . So, $F=E$ .
$\implies AE=2BE$

This post has been edited 1 time. Last edited by thegreatp.d, Jul 10, 2019, 5:02 PM
Reason: Typo

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#4 h Jul 10, 2019, 3:55 PM • 2 Y

Y by Adventure10, Mango247

We have: $0 = \overrightarrow{AD} . \overrightarrow{CE} = \dfrac{1}{2} (\overrightarrow{AB} + \overrightarrow{AC})(\overrightarrow{CB} + \overrightarrow{BE}) = \dfrac{1}{2} (\overrightarrow{AB} . \overrightarrow{CB} + \overrightarrow{AB} . \overrightarrow{BE} + \overrightarrow{AC} . \overrightarrow{CB} + \overrightarrow{CB} . \overrightarrow{BE})$ $= \dfrac{1}{2} \left(BC^2 - AB . BE - \dfrac{BC . BE}{\sqrt{2}}\right)$
Then: $BE = \dfrac{BC^2 \sqrt{2}}{AB \sqrt{2} + BC} = \dfrac{BC^2 \sqrt{2}}{3 BC} = \dfrac{BC \sqrt{2}}{3} = \dfrac{AB}{3}$ or $AE = 2 BE$

This post has been edited 1 time. Last edited by khanhnx, Jul 10, 2019, 3:56 PM

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#5 h Jul 11, 2019, 3:11 PM • 1 Y

Y by Adventure10

Let $M$ be the midpoint of $AB$ and $G$ the intersection point of $AD, CM.$ Obviously $G$ is the

centroid of $ABC$ and the orthocenter of $CAE.$ Hence, $EG||BD$ and $\frac{{AE}}{{EB}} = \frac{{AG}}{{GD}} = 2$

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Reason: typo

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#6 h Oct 30, 2023, 1:15 PM

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$F$ - reflection of $C$ about $AB$ , $\{\ G\ \}\in CE\cap BF$ , then $E$ is centroid of $\triangle BCF$ ( $G$ is midpoint of $BF$ ), done.

Best regards,
sunken rock

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#7 h May 17, 2024, 6:02 AM

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centroid + orthocenter solution, with an auxiliary point

This post has been edited 5 times. Last edited by parmenides51, Jun 4, 2024, 9:36 AM

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