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The next stage of mathematics learning in the seventh grade of People's Education Press refers to the triangle.
1is ∠BPD∠2 is ∠CPG.

It is proved that ≈ 1 is the external angle of △ABP.

∴∠ 1=∠BAP+∠ABP

What's Ad, be and cf? Angle bisector of △ABC

∴∠BAP+∠ABP+∠DCG= 180 /2=90

∴∠ 1+∠DCG=90

∵PG⊥BC in G

∴∠2++∠DCG=90

∴∠ 1=∠2?

Is that clear? Let me know if you have any questions.

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