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Both the triangle ABC and the triangle ECD are isosceles right triangles. The angle ACB = angle DCE = 90 ° D is a point on the edge of AB and connected with Ae. When de = 17 and AE = 15, the triangle ABC and ECD are isosceles right triangles, Find the length of ad
D is a point on the edge of AB, connecting AE, ∵ triangle ABC and triangle ECD are isosceles right triangles, ∵ ACB = ∠ DCE = 90 degrees, BAC = 45 degrees = ∵ Dec a.d.c.e four points are circular ? DCE = 90 ° ? De is the diameter of the circle, ? DAE = 90 ° AED is a right triangle, ad ∵ AE ? De is a right triangle
Both △ ABC and △ ECD are isosceles right triangles, ∠ ACB = ∠ DCE = 90 ° and D is a point on AB side. It is proved that BD = AE
? ABC, △ ECD is isosceles right angle, AC = BC, EC = DC
And ? ACB = ∠ DCE = 90 °, ACB - ∠ ACD = ∠ DCE - ∠ ACD
That is ∠ ace = ∠ BCD
∴△AEC≌△BDC(SAS)
∴AE=BD
It is known that △ ABC and △ ECD are isosceles right triangles, ∠ ACB = ∠ DCE = 90, D is a point on AB, and it is proved that BD = AE
? ABC, △ ECD is isosceles right angle, AC = BC, EC = DC
And ? ACB = ∠ DCE = 90 °, ACB - ∠ ACD = ∠ DCE - ∠ ACD
That is ∠ ace = ∠ BCD
∴△AEC≌△BDC(SAS)
∴AE=BD
As shown in the figure, both triangle ABC and triangle ECD are isosceles right triangles, angle ACB = angle DCE = 90 ° and D is a point on AB side
It is proved that: the connection of be
As shown in the figure, △ ABC and △ ECD are isosceles right triangles, ∠ ACB = ∠ DCE = 90 ° D, D is a point on the edge of AB to prove AE = BD
∵∠ECD=∠ACB=90º
∴∠ECD-∠ACD=∠ACB-∠ACD
That is ∠ ECA = ∠ DCB
And ∵ CE = CD, CA = CB
∴ΔACE≌ΔBCD
∴BD=AE
The angle formed by one side of a triangle and the opposite extension of the other side is called the outer angle of the triangle. As shown in the figure, ∠ ACD is an external angle of △ ABC, in which ∠ ACD is not adjacent. What is the equivalent relationship between foot a, teach B and angle ACD
∠ACD=∠A+∠B
The reasons are as follows.
∵∠A+∠B+∠ACB=180
∠ACD+∠ACB=180
∴∠ACD=∠A+∠B
Language expression: any exterior angle of a triangle is equal to the sum of two interior angles which are not adjacent to it
In △ ABC, ab = AC, ad is the bisector of the exterior angle of △ ABC. It is known that ∠ BAC = ∠ ACD to prove △ ABC congruence △ CDA
Proof: take point E on Ba extension line
∵AB＝AC
∴∠B＝∠ACB
∴∠CAE＝∠B+∠ACB＝2∠ACB
∵ ad bisection ∵ CAE
∴∠CAD＝∠CAE/2＝∠ACB
∵∠BAC＝∠ACD
∴△ABC≌△CDA （ASA）
As shown in the figure, △ ABC is a steel frame, ab = AC, ad is the bracket connecting point D of a and BC Verification: △ abd ≌ △ ACD
Prove that: ∵ D is the midpoint of BC,
∴BD=DC．
In △ abd and △ ACD,
A kind of
AB＝AC
BD＝CD
AD＝AD ，
∴△ABD≌△ACD（SSS）．
As shown in the figure, △ ABC is a steel frame, ab = AC, ad is the bracket connecting point D of a and BC Verification: △ abd ≌ △ ACD
Prove that: ∵ D is the midpoint of BC,
∴BD=DC．
In △ abd and △ ACD,
A kind of
AB＝AC
BD＝CD
AD＝AD ，
∴△ABD≌△ACD（SSS）．
As shown in the figure, △ ABC is a steel frame, ab = AC, ad is the bracket connecting point D of a and BC Verification: △ abd ≌ △ ACD
Prove that: ∵ D is the midpoint of BC,
∴BD=DC．
In △ abd and △ ACD,
A kind of
AB＝AC
BD＝CD
AD＝AD ，
∴△ABD≌△ACD（SSS）．
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