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Nine-grade mathematical circle geometry problems
1. Connect the advertisement. Because of tangency, AD is perpendicular to BC. So △ADB≌ Delta ADC. So BD=DC.

BE = CF because AB=AC and AE=AF (radius of the same circle) are known.

∠B=∠C

So △ bed △ CFD

So DE=DF

2. Because BC=8, BD = DC = 4;; AD = AE = AF = 3; According to Pythagorean Theorem, AB=AC=5.

Because ∠A is public, AE=AF, AB=AC, so △AEF∽△ABC, so AE: AB = EF: BC.

That is, 3: 5 = ef: 8, and EF=*