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The sixth problem on page 52 of the first volume of eighth grade mathematics of People's Education Press.
Because ad bisects the angle bac De perpendicular to ab and the angle df perpendicular to ac. Known. So ... De equals the property of df angle bisector. Ad is equal to the ad in the angle of rt3 aed rt3 AFD. De is equal to df, so the angle Aed of rt3 is identical to the angle afd of rt3. HL。 So ae equals af. Because ad divides BAC into two. Therefore, the three lines of the ad vertical EF isosceles triangle are integrated.