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Source: Tuymaada 2016. junior league\ P2

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#1 h Jul 22, 2016, 10:02 AM • 1 Y

Y by Adventure10

The point $D$ on the altitude $AA_1$ of an acute triangle $ABC$ is such that
$\angle BDC=90^\circ$ ; $H$ is the orthocentre of $ABC$ . A circle
with diameter $AH$ is constructed. Prove that the tangent drawn from $B$
to this circle is equal to $BD$ .

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#2 h Jul 22, 2016, 11:54 AM • 1 Y

Y by Adventure10

Let $C_1\equiv CH\cap AB$ $\Longrightarrow$ $C_1\in \odot (BDC)$ and $C_1\in \odot (ACA_1)$ $\Longrightarrow$ $\measuredangle C_1DB=\measuredangle C_1CB=\measuredangle C_1AA_1$ $\Longrightarrow$ $BD$ is tangent to $\odot (ADC_1)$ $\Longrightarrow$ $BD^2=BC_1.BA...(1)$ . Let $X\in \odot (AHC_1)$ such that $BX$ is tangent to $\odot (AHC_1)$ $\Longrightarrow$ $BX^2$ $=$ $BC_1.BA$ $...(2)$ $\Longrightarrow$ by $(1)$ and $(2)$ we get $BX=BD$ .

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#3 h Aug 28, 2018, 7:23 PM • 2 Y

Y by Adventure10, Mango247

see figure:

Attachments:

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2125 posts

#4 h Jul 27, 2019, 1:54 PM • 2 Y

Y by Adventure10, Mango247

Problem: Let $\Delta DEF$ be the orthic triangle WRT $\Delta ABC$ with orthocenter $H$ . Let $M$ be midpoint of $\overline{BC}$ . Let $N$ $\equiv$ $\odot (BFEC)$ $\cap$ $\overline{AD}$ . Let $BL,BK$ be tangents from $B$ to $\odot (AEF)$ . Show, $BK=BN=BL$

Proof: Let $N' \in \overline{AD}$ , such $BK=BN'=BL$ . Perform inversion $\Psi$ around $\odot (KLN')$ . Since, $\Psi (F)$ $\equiv$ $A$ and $\Psi ( E)$ $\equiv$ $H$ $\implies$ $\Psi ( \odot (BFEC))$ $\equiv$ $AH$ $\implies$ $N' \in \odot (BFEC)$ $\implies$ $N \equiv N'$

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primesarespecial

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primesarespecial

#5 h Jul 3, 2021, 4:08 PM

Y by

Hehe,this was so simple and direct,but still I will write the solution. :blush:

:blush:

From similarity ,we have $BD^2=BA_{1}.BC$ .
From power of $B$ w.r.t $\odot(AC_{1}A_{1}C)$
$BA_{1}.BC=BC_{1}.BA=BD^2$ .
Hence we are done.

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