1. As shown in Figure 4,△ABC has an incircle ⊙O1 that touches side BC at point D,⊙O2 is the excircle of △ABC inside ∠A. If O1B=6,O1C=3,O1D=2, then O1O2=
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Solution
⇒⇒⇒⇒∠O1BO2=∠O1CO2=90∘O1、C、O2、B are concyclic ∠O1CD=∠O1O2B△O1DC∽△O1BO2O1O2=O1DO1B⋅O1C=23×6=9.
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Source: NuminaMath-1.5,
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