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As shown in the figure, in isosceles RT △ ABC, AC = BC, e is a point on the extension line of BC, connecting AE, if the vertical line of AE intersects the bisector of ACB at P, AC intersects F (2) Try to judge the quantitative relationship among BC, CE and CP (can you write it in the way of grade two) (3) If BC = 7, when CE = ----, AF = ef (fill in the conclusion directly)
Upstairs obviously copied my original answer!
Sorry, I don't know how to use the knowledge of grade two~
But at that time, after I answered, he said that their teacher had already said that maybe you are all sophomores ~ I suggest you ask him!
See the following link:
In isosceles △ ABC, ab = AC, D is the interior point of △ ABC, ∠ ADB = ∠ ADC
Prove that: make ∠ CAE = ∠ bad, make AE = ad (point E and D are on both sides of AC). Connect CE and de
∵AC=AB,AE=AD,∠CAE=∠BAD.
∴⊿CAE≌⊿BAD(SAS),CE=BD;------------------------------(1)
Moreover, AEC = ADB = ADC
∵AE=AD.
Then, AEC - AED = ADC - ADE
That is, CE = CD
Therefore, CD = BD, ∠ DBC = DCB
In △ ABC, ab = AC, D is the midpoint of AB, extend AB to e, make be = AB, indicating △ ADC ～△ ace
Let AB = AC = be = a,
There are ad = A / 2, AE = 2A,
In △ ADC and △ ace,
AD=a/2,AC=a,
AC=a,AE=2a,
∴AD：AC=AC：AE=1：2,
A is the common angle,
A set of corresponding angles are equal, and the two sides of the angle are proportional, and the two triangles are similar
As shown in the figure, ∠ BAC = 90 °, ab = AC, point D is on AC, point E is on the extension line of Ba, BD = CE, and the extension line of BD intersects CE at F. try to prove that: BF ⊥ CE
In RT △ bad and RT △ CAE, BD = ceab = AC ≌ RT △ bad ≌ RT △ CAE (HL), ≌ abd = ≌ ace, ≌ ADB = ≌ CDF, ∵ abd + ≌ ADF = ≌ ACE + ≌ CDF, and ≂ abd + ≌ ADF = 90 degree
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