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Point $T$ is inside the square $ABCD$. Let $A_1,B_1,C_1,D_1$ the other intersection point of the lines $AT,BT,CT,DT$ respectively and the circumcircle of the square $ABCD$. Prove: $$|A_1B_1|\cdot|C_1D_1|=|A_1D_1|\cdot|B_1C_1|$$

My attempt:

I was looking for the inscribed angles of the same measure: $$\measuredangle ABB_1=\measuredangle AA_1B_1\;\&\;\measuredangle BTA=\measuredangle B_1TA_1\implies\;\Delta ABT\;{\sim}\;\Delta A_1B_1T$$ Analogously:

$$\Delta TAD_1{\sim}\Delta TA_1D$$$$\;\Delta C_1D_1T\;{\sim}\Delta CDT$$$$\Delta D_1A_1T{\sim}\Delta DAT$$$$\Delta B_1C_1T{\sim}\Delta CBT$$

Also, $\measuredangle DB_1B=\measuredangle BA_1D$, so $DB_1B$ and $BA_1D$ are right triangles.

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However, I couldn't find any triangles with useful information. May I ask for advice on solving the problem? Thank you in advance!

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1 Answer 1

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Because $$\frac{A_1D_1\cdot B_1C_1}{A_1B_1\cdot C_1D_1}=\frac{\frac{A_1D_1}{AD}\cdot\frac{B_1C_1}{BC}}{\frac{A_1B_1}{AB}\cdot\frac{C_1D_1}{CD}}=\frac{\frac{A_1T}{DT}\cdot\frac{C_1T}{BT}}{\frac{A_1T}{BT}\cdot\frac{C_1T}{DT}}=1$$

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