Tam giác ABC vuông ở A có AB<AC, trung tuyến AD. Kẻ đường thẳng vuông góc với AD, cắt AC,AB tại E,F. Đường cao AH của tam giác ABC cắt EF tại I. Chứng minh $\frac{S_{ABC}}{S{AEF}}$=( $\frac{AD}{AI}$ ) ² Giúp mình với mình đang cần gấp

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