Circle through A of parallelogram ABCD

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Circle through A of parallelogram ABCD

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WakeUp

#1 h Nov 17, 2010, 7:07 PM • 2 Y

Y by Adventure10, Mango247

Given a parallelogram $ABCD$ . A circle passing through $A$ meets the line segments $AB, AC$ and $AD$ at inner points $M,K,N$ , respectively. Prove that
$|AB|\cdot |AM | + |AD|\cdot |AN|=|AK|\cdot |AC|$

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#2 h Nov 18, 2010, 12:25 AM • 1 Y

Y by Adventure10

Let $\omega$ be the object circle and $\tau$ the perpendicular line to $AC$ through $C.$ $\tau$ is the inverse of $\omega$ under the inversion with pole $A$ and power $\overline{AK }\cdot \overline{AC}$ $\Longrightarrow$ $M' \equiv AB \cap \tau$ and $N' \equiv AD \cap \tau$ are the inverses of $M,N.$ Thus $\overline{AK} \cdot \overline{AC}=\overline{AM} \cdot \overline{AM'} =\overline{AN} \cdot \overline{AN'}.$ But from $\triangle M'BC \sim \triangle M'AN' \sim \triangle CDN',$ we deduce that

$\frac{\overline{AB}}{\overline{AM'}}+\frac{\overline{AD}}{\overline{AN'}}=1 \Longrightarrow \overline{AB} \cdot \frac{\overline{AM}}{ \overline{AK} \cdot \overline{AC}}+\overline{AD} \cdot \frac{\overline{AN}}{\overline{AK} \cdot \overline{AC}}=1$

$\Longrightarrow \overline{AB} \cdot \overline{AM}+\overline{AD} \cdot \overline{AN}=\overline{AK} \cdot \overline{AC}.$

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#3 h Nov 19, 2010, 12:38 AM • 2 Y

Y by Adventure10, Mango247

By Ptolemy: $AM*NK+AN*KM = AK*NM$ (1)

Also $\angle{KMN}=\angle{KAN}=\angle{ACB}$ and $\angle{MNK}=\angle{MAK}$ so $ACB$ is similar to $NMK$ and therefore:

$\frac{AC}{MN}=\frac{AB}{NK}=\frac{BC}{MK}=\frac{AD}{MK}= k$

which implies, $NK= AB*\frac{1}{k}$ , $KM = AD*\frac{1}{k}$ and $NM= AC*\frac{1}{k}$ . Substituting this into (1) we get the desired result.

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ririri

#4 h Apr 17, 2019, 5:10 PM • 2 Y

Y by Adventure10, Nari_Tom

Place the origin at $A$ . It is well known that the function $\pi(X) = \operatorname{Pow}_A(X) - \operatorname{Pow}_{(KMN)}(X)$ is linear. Consequently,
$\begin{align*} AC\cdot AK &= CA^2 - CA\cdot CK \\ &= \pi(C) \\ &= \pi(B+D) \\ &= \pi(B) + \pi(D) \\ &= AB\cdot AM + AD\cdot AN. \end{align*}$

This post has been edited 1 time. Last edited by ririri, Apr 20, 2019, 2:49 PM
Reason: wording

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#5 h Apr 18, 2019, 10:06 AM • 1 Y

Y by Adventure10

If the circle $\odot(MBK)$ intersects $AC$ second time at $P$ and circle $\odot(DNK)$ intersects $AC$ second time at $Q$ , it's not big deal to prove $BP\parallel DQ$ , wherefrom $PC=AQ\ (\ 1 \ )$ and applying p.o.p. we get $AN\cdot AD=AK\cdot AQ\ (\ 2\ )$ and $AM\cdot AB=AK\cdot AP\ (\ 3\ )$ . Adding $(2)$ and $(3)$ side by side and taking into account $(1)$ , we get the desired relation.

Best regards,
sunken rock

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#6 h May 24, 2024, 11:16 AM

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Claim: $\Delta MKN \sim \Delta CBA$
Proof: Angle chase.

Now from Ptolemy's theorem, $$AM \cdot NK + AN \cdot MK = AK \cdot MN$. Finding $MN, NK, MK$ wrt $AB,BC,CA$ and substituting gives the result. $\square$

This post has been edited 1 time. Last edited by AshAuktober, May 24, 2024, 11:16 AM

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