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Geometry Difficulty 5.6 AIME, harder Prove it

II. (25 points) As shown in Figure 1, it is known that ⊙O1\odot O_{1} passes through two vertices AA and BB of trapezoid ABCDABCD and is tangent to the side CDCD at NN; ⊙O2\odot O_{2} passes through points CC and DD and is tangent to the side ABAB at point MM. Prove:
AM⋅MB=CN⋅ND A M \cdot M B = C N \cdot N D

Solution

Since AD∥BC,∴OBOA=OCOD=t2. \begin{array}{l} \text{Since } AD \parallel BC, \\ \therefore \frac{OB}{OA}=\frac{OC}{OD}=t^{2} . \end{array}

Thus, OB=OA⋅t2=at2OB = OA \cdot t^{2} = a t^{2},
OC=OD⋅t2=bt2.  OC = OD \cdot t^{2} = b t^{2} \text{. }

Since ONON is tangent to ⊙◯1\odot \bigcirc_{1},
we have ON2=OA⋅OB=a2t2. ∴ON=at. ∴CN=OC−ON=bt2−at, ND=ON−OD=at−b. Thus, CN⋅ND=(bt2−at)(at−b)=(bt−a)(at−b)t. \begin{array}{l} \text{we have } ON^{2} = OA \cdot OB \\ = a^{2} t^{2} \text{. } \\ \therefore ON = a t \text{. } \\ \therefore CN = OC - ON = b t^{2} - a t \text{, } \\ ND = ON - OD = a t - b \text{. } \\ \text{Thus, } CN \cdot ND = (b t^{2} - a t)(a t - b) \\ = (b t - a)(a t - b) t . \\ \end{array}

Similarly, AM=bt−aAM = b t - a, MB=at2−btMB = a t^{2} - b t.
∴AM⋅MB=(bt−a)(at2−bt)=(bt−a)(at−b)t. \begin{aligned} \therefore AM \cdot MB & = (b t - a)(a t^{2} - b t) \\ & = (b t - a)(a t - b) t . \end{aligned}

Therefore, AM⋅MB=CN⋅NDAM \cdot MB = CN \cdot ND.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.