AH = HE wanted, starting with isosceles, projections

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AH = HE wanted, starting with isosceles, projections

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Source: 2016 France JBMO TST 1.1

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#1 h Jul 20, 2020, 4:45 AM

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Let $ABC$ be an isosceles triangle with vertex in $A$ , with $\angle A '\ne 90^o$ . Let $D$ be the point on $BC$ such that $AD \perp AB$ . Let $E$ be the projection of $D$ on the $AC$ . Finally, let $H$ be the middle of $[BC]$ . Prove that $AH = HE$ .

This post has been edited 1 time. Last edited by parmenides51, Jul 20, 2020, 4:58 AM
Reason: typo, <A \ne 90^o instead of <A = 90^o

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#2 h Jul 20, 2020, 4:53 AM • 2 Y

Y by Mango247, Mango247

Wait, first two sentences are possible only if $D=C$ , am i right?
Or you have typo?

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#3 h Jul 20, 2020, 4:57 AM

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typo, $\angle A \ne 90^o$

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#5 h Jul 20, 2020, 8:30 AM

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parmenides51 wrote:

Let $ABC$ be an isosceles triangle with vertex in $A$ , with $\angle A '\ne 90^o$ . Let $D$ be the point on $BC$ such that $AD \perp AB$ . Let $E$ be the projection of $D$ on the $AC$ . Finally, let $H$ be the middle of $[BC]$ . Prove that $AH = HE$ .

Note that $AH \perp BC$ , so: $\angle AHD= \angle AED=90^o$ thus $AHED$ is cyclic, which means: $\angle HEA=\angle HDA= 90^o-\angle ABC= 90^o-\angle HCA=\angle HAC$ hence $\angle HEA=\angle HAC \implies AH=HE$ .
$\square$

This post has been edited 3 times. Last edited by CROWmatician, Jul 20, 2020, 8:41 AM
Reason: .

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