In the triangle ABC, D is the point above AB, e is the point above AC, ∠ ACD = ∠ B, ad square = AE · AC

In the triangle ABC, D is the point above AB, e is the point above AC, ∠ ACD = ∠ B, ad square = AE · AC

In △ ACD and △ ABC, ∵ ACD = B, ∵ CAD = BAC, ∵ CAD ∽ BAC, ∵ ad ∶ AC = AC ∶ AB, ∵ AD & # 178; = AE · AC, ∵ ad ∶ AC = AE ∶ ad, ∵ AE ∶ ad = AC ∶ AB, ∵ AE ∶ AC = ad ∶ AB, ∵ de / / BC
The solution set of known inequality ax + 1 > x + A is X
ax+1>x+a
(a-1)x>a-1
Because the solution set is X
Original formula:
ax-x>a-1
(a-1)x>a-1
Because when x1, x1, the equation doesn't hold
When a
As shown in the figure, e and D are two points on the edge of BC in △ ABC, ad = AE. To prove △ Abe ≌ △ ACD, supplementary conditions should be added______ .
When BD = Ce (be = CD can be obtained) or AB = AC (be = C can be obtained) or ∠ B = ∠ C or ∠ BAE = ∠ CAD, ≌ Abe ≌ ACD. Therefore, ab = AC or ∠ B = ∠ C or ∠ BAE = ∠ CAD
If the inequality AXA is 1 / 2, then the value range of a is——————
If the direction of inequality AXA changes, then a
As shown in the figure, CE and CB are the midlines of △ ABC and △ ADC respectively, and ab = AC. verification: CD = 2ce
It is proved that: extend CE to F, make ef = CE, connect FB. ∵ CE is the middle line of △ ABC, ∵ AE = EB, and ∵ ∵ AEC = ∠ bef, ≌ AEC ≌ bef, (SAS) ∵ a = ∠ EBF, AC = FB. ∵ AB = AC,