right angle inside a triangle, symmetrical of a midpoint wrt a height related

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right angle inside a triangle, symmetrical of a midpoint wrt a height related

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#1 h Sep 16, 2018, 8:24 PM • 2 Y

Y by Adventure10, Mango247

Altitudes $AD$ and $CE$ of an acute angled triangle $ABC$ intersect at point $H$ . Let $K$ be the midpoint of side $AC$ and $P$ be the midpoint of segment $DE$ . Let $Q$ be the symmetrical point of $K$ wrt line $AD$ . Prove that $\angle QPH = 90^o$

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#2 h Sep 18, 2018, 2:04 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

$AQDK$ is a rhombus; let $F=AD\cap KQ, R$ - reflection of $Q$ in $P$ , hence $PF$ is midline of triangles $ADE$ and $KQR$ , thus $KR\parallel =AE$ , hence $R$ is the reflection of $K$ in $CE$ , therefore $RH=KH=QH\implies PQ\bot PH$ .

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#3 h Sep 18, 2018, 2:37 PM • 2 Y

Y by Adventure10, Mango247

How did you conclude that R is the reflection of K in CE? I specifically didn't understand how are K and R equidistant from CE?

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#4 h Sep 19, 2018, 9:42 AM • 2 Y

Y by Adventure10, Mango247

sai_sanjeev wrote:

How did you conclude that R is the reflection of K in CE? I specifically didn't understand how are K and R equidistant from CE?

Since $KR \parallel AE$ , we have that $AERK$ is a parallelogram. But then $ER$ and $KC$ are parallel as well and have the same length, so $ERKC$ is a parallelogram. On the other hand, $KE=KC$ so this is even a rhombus and the claim follows. I hope that helps.

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#5 h Jul 17, 2020, 11:08 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $a,b,c$ be complex numbers lying on unit circle centered at $0$ such that they represent points $A,B,C$ respectively on Argand plane. We have $2k=a+c,\ h=a+b+c,\ 2d=a+b+c-\frac{bc}{a},\ 2e=a+b+c-\frac{ab}{c}, 2p=d+e=a+b+c-b\cdot\frac{c^2+a^2}{2ac}.$ Line $AH$ crosses the circumcircle for the second time in point with coordinate $-bc/a$ . From reflection of point formula $q=a-\frac{bc}{a}+bc\cdot\overline{k}=a-\frac{bc}{a}+bc\cdot\frac{a+c}{2ac}=\frac{2a^2+ab-bc}{2a}.$ Auxiliary calculations
$q-p=\frac{4a^2c+2abc-2bc^2-2ac(a+b+c)+b(a^2+c^2)}{4ac}=\frac{(a-c)(ab+bc+2ac)}{4ac},$ $h-p=\frac{(a+c)(ab+bc+2ac)}{4ac}.$ Hence
$\frac{q-p}{h-p}=\frac{a-c}{a+c}=-\overline{\left(\frac{a-c}{a+c}\right)}=-\overline{\left(\frac{q-p}{h-p}\right)}$ As we know
$\angle QPH = 90^o\iff \frac{q-p}{h-p}=-\overline{\left(\frac{q-p}{h-p}\right)}.$ QED

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