RT triangle ABC, ∠ a = 90 °, bisector of ∠ B intersects AC at D, vertical line from a as BC intersects BD at e, and vertical line from D as DF, so aefd is proved to be diamond

From the angle bisector theorem, we can get ad = DF, ∠ abd = ∠ CBD, let the point of AE intersection BC be g, because Ag ⊥ BC, DF ⊥ BC, so Ag / / DF, because ⊥ DBG + ∠ beg = 90 °, and ⊥ abd + ∠ BDA = 90 °, so ⊥ BDA = ∠ beg = ∠ AED, so AE = ad = DF, and DF / / AE, so quadrilateral aefd is parallelogram, and ad = DF

RT triangle ABC, ∠ a = 90 °, bisector of ∠ B intersects AC at D, vertical line from a as BC intersects BD at e, and vertical line from D as DF, so aefd is proved to be diamond

RT triangle ABC, ∠ a = 90 °, bisector of ∠ B intersects AC at D, vertical line from a as BC intersects BD at e, and vertical line from D as DF, so aefd is proved to be diamond

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