Picture Below! Given: Isosceles Triangle ABC with AD = AB Prove: Triangle BDA ~ Triangle ABC
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I'm guessing the ~ means similar. Anyway I can't see how the 2 triangles can be similar.
The thing with the proof it's not like by seeing is by knowing the facts. I'm trying to see if theres something that goes with this.
In order to be similar, the angles of each triangle must equal the angles of the other. I can't see how they equal each other.
Soo what we know is that angle A and B are = and that AD ='s AB
If a line divides two sides of a triangle proportionally, then it is parallel to the third side
Specifically angle DAB = angle B
Well. DAB + CAD right?
Where is the line that divides two sides of the triangle. and by DAB + CAD do you mean DAB = CAD?
|dw:1393312074782:dw|
DAB alone cant = angle B
Is there any additional information about this problem?
No, but In a step we may need to draw something
It says we are given isosceles triangle ABC and that AD = AB. That is not expressed properly.
Don't look at the picture to what's given, I didn't draw it perfectly lol. AD = AB
"If a line divides two sides of a triangle proportionally, then it is parallel to the third side" means: |dw:1393312450627:dw|
|dw:1393312470195:dw|
So it is definite that angle CAD = angle DAB?
No, but If AD and AB are equal, that creates another Isosceles ?
|dw:1393312634430:dw| Thats the picture the book gives for "if a line divides two sides of a triangle proportionally, then it is parallel to the third
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