1, as shown in the figure: connecting OB, OC, OD, OD and BC intersect at point P, because OB, OC, OD are the radius of the circle and DE is the tangent of the circle, so OB=OC, OD is perpendicular to DE, from which it can be concluded that △BOC is an isosceles triangle, and because DE∑BC and OP are perpendicular to BC, OP is isosceles △BOC.

Is this available? Arc BD = arc CD, because if the angles of two circles inscribed with a triangle correspond to equal arc segments, then the two angles are also equal, so ∠BAD=∠DAE, ∠ACB=∠ADB (corresponding to the same arc AB), because DE∨BC∠ACB =∞.

∠AED=∠ADB, ∠BAD=∠DAE, so △ Abd △ aed.

2. Because if the arc segments corresponding to the angles of two circles inscribed in a triangle are equal, then the two angles are also equal, so ∠ ADC = ∠ ABC = 45 (corresponding to the same arc AC), since FA is known to be perpendicular to DA and AF=AD(△DAF is an isosceles right triangle), S △ DAF = 1. Because △ABD∽△ADE, AB/AD=AD/AE.

So we can get AB*AE=AD? Because S△ABE= 1/2*AB*AEsin∠BAE (triangle area formula),

Available s △ Abe =1/2 * ab * aesin ∠ BAE =1/2 * ad? Sin∠BAE, because it is known that △ABC is an acute triangle, △ BAE is less than 90, and SIN ∠BAE is less than 1, so 1/2*AD? sin∠BAES & lt； 1/2AD？ , so △ Abe < S? △DAF