According to the problem, in triangle and triangle, angle EAD is equal to angle FAD, angle AED is equal to angle AFD, and AD is equal to AD, so these two triangles are congruent! As can be seen from the above, in triangle and triangle, angle bed equals angle CFD equals 90 degrees, ED equals FD and BD equals CD, so these two triangles are also congruent, so BD equals CF Note: When proving congruence, it must be stated that angle AED equals angle AFD equals 90 degrees, and angle bed equals angle CFD equals 90 degrees, otherwise congruence cannot be stated, because when angle bed is acute, there is no need to draw two line segments BD (or CD) that meet the requirements. College entrance examination questions can never be solved by pure theorems. It is most important to prove that your method is correct in your own way! As for the corner, I don't remember it clearly! But you should pay attention, you have to prove that congruent triangles is a right triangle! In particular, I didn't understand what you wrote, but your first sentence must be wrong if there are no conditions! Think about it, in a triangle, if two corresponding angles are equal, does it mean that three angles are equal, and three angles are equal to prove that two triangles are similar, because the corresponding hypotenuses are equal, can you prove congruence?

8. Because AD is perpendicular to CB and DB is perpendicular to CB, angle DBC= angle ACB=90 degrees. In RT triangle ABC and RT triangle DCB, CB=BC AB=DC, so it is congruent and the angle is equal to 10. Congruences (SAS) with the same vertex angle and the same two sides are then parallel through inner angles. 1 1。 Because FB=CE and FB FC=CE FC, BC=EF. Because parallel angles are equal, congruence is obtained, so edges are equal.

Angle CBD = BDA；; ; So BF = df, af = cf in the right triangle CDF, CD=3, cfdf = 9；; And df-cf = 9；; (CF DF)*(DF-CF)= 9； Then df-cf =1; Solvable: DF=5, cf = 4；; Then the BDF area of the triangle =DF*AB/2=5*3/2=7.5.