right angled triangle: does BD bisects angle <ABC?

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defeated_soldier

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I have a right angled triangle ABC

<B is the right angle.

now i draw BD as an altitude on the hypotenuse .

does BD bisects angle <ABC ?
 
Re: right angled triangle

defeated_soldier said:
I have a right angled triangle ABC

<B is the right angle.

now i draw BD as an altitude on the hypotenuse .

does BD bisects angle <ABC ?
Click to expand...

IF the right triangle happens to be isosceles (that is, if AB = BC), then the altitude to the hypotenuse also bisects the angle at B. But in general, one cannot assume that the altitude is also an angle bisector.

It is always true that the altitude to the hypotenuse in any right triangle divides the original triangle into two smaller right triangles which are similar to the original triangle and to each other.
 
Re: right angled triangle

Mrspi said:
It is always true that the altitude to the hypotenuse in any right triangle divides the original triangle into two smaller right triangles which are similar to the original triangle and to each other.
Click to expand...
yes .but i cant prove it . and thats why i asked the question , because if the altitude bisects then i can use " AA" test to prove similarity .

SSS test, SAS does also fails to prove similarity .

I am now worried about the similarity . but my book says they are similar ...i dont know how they proe that .

can we prove two smaller triangles are similar ?(assuming AB!=BC)
 
defeated_soldier said:
yes .but i cant prove it .
Click to expand...
You can't prove which? That the sub-triangles, which have to be right (since the altitude is, by definition, perpendicular to the opposite side), are similar? Try using the fact that they are right, and that they share one of the angles. Then, by what you know about the angle-sum of any triangle, you know that the measure of the third angle is the same. And what do you know about the similarity of any two triangles which share the same three angle measures?

But that similarity says nothing about the alititude bisecting the other side. In general, it won't.

Eliz.
 
stapel said:
defeated_soldier said:
yes .but i cant prove it .
Click to expand...
Try using the fact that they are right, and that they share one of the angles.

Eliz.
Click to expand...

no , they dont share .

I am saying two triangles i.e ABD and BDC. [ BD is the altitude , B is right angle]
they dont share any common angle.


If you say , ABC vs ABD or ABC vs BDC ....i agree., in that case they will share <A or <C respectively.


But , is not the two sub triangles ABD and BDC are similar . i wanted to prove this ....and i am stuck here.
 
defeated_soldier said:
no , they dont share .

I am saying two triangles i.e ABD and BDC. [ BD is the altitude , B is right angle]
they dont share any common angle.


If you say , ABC vs ABD or ABC vs BDC ....i agree., in that case they will share <A or <C respectively.


But , is not the two sub triangles ABD and BDC are similar . i wanted to prove this ....and i am stuck here.
Click to expand...

You are very close.

triangle ABC ~ triangle ADB (be sure you name similar triangles so that the equal angles are named in the same relative positions in the names)

triangle ABC ~ triangle BDC

Now, since each of the sub-triangles is similar to the original triangle ABC, by the transitive property, they must be similar to each other:

triangle ADB ~ triangle BDC
 
you used the transitive property . ok fine....i agree.

but one thing i dont understand , if the two triangles are similar really ...then they must follow any one of AA,SSS,SAS tests ....then why we have to follow the other way (i.e transitive way thoug its valid one) .............why there is no direct proof ?

ALL i know is , if 2 triangles are similar then they must obey any one of the tests

(i)AA
(ii)SSS
(iii)SAS

As I knew , triangles are similar so they should obey the above !

whats wrong with my approach ? why its not working .




Another question is , i am bit curious to know whether the altitude bisects the hypotenuse ?


Thanks for the response.
 
stapel said:
Try using the fact...they share one of the angles.
Click to expand...
defeated_soldier said:
no , they dont share .
Click to expand...
You said that you have a right triangle, ABC, with the right angle at B. You have drawn an line inside angle ABC to side AC, and this line is the altitude from B to AC, meeting AC at point D.

The segment BD splits triangle ABC into two (right) triangles, ABD (with right angle BDA) and BCD (with right angle CDB). Draw this.

How do you get that triangle ABC and triangle ABD do not share angle BAD?

How do you get that triangle ABC and triangle BCD do not share angle BCD?

Please be specific.

defeated_soldier said:
i am bit curious to know whether the altitude bisects the hypotenuse ?
Click to expand...
At least two replies have said, "No". Since these did not answer your question, it might be helpful if you clarified what you are asking.

Thank you.

Eliz.
 
Code: 
B

a              c

C           b              A
defeated_soldier (wish you'd change that to Mathsoldier; it depresses me...)
please start using the STANDARD lettering for right triangles as shown above:
across A is BC, labelled a
across B is AC, labelled b
and the hypotenuse is across the 90degree C, labelled c.
Using these standards makes it easier for EVERYONE.

Code: 
B
       D
              

C                          A

The height to the hypotenuse is line CD: CD is perpendicular to AB,
so triangles ACD and BCD are both right triangles.

If we let angleABC = x, then angleBAC = 90 - x.

Therefore, in triangle ACD, angleACD = x
and in triangle BCD, angle BCD = 90 - x

So all 3 triangles have exactly the same angle size.

That's all you need as PROOF.
 
Denis said:
Code: 
B

a              c

C           b              A
Click to expand...
But didn't the student specify that the right angle is at vertex B, not vertex C...?

Eliz.
 
YES, Stapel; my suggestion has nothing to do with what the student DID;
I'm suggesting that in the FUTURE, he uses C instead of B; c'est tout :wink:
 
Denis said:
YES, Stapel; my suggestion has nothing to do with what the student DID;
I'm suggesting that in the FUTURE, he uses C instead of B; c'est tout :wink:
Click to expand...

Ok, dumb question Denis...but WHY????
 
Denis said:
defeated_soldier (wish you'd change that to Mathsoldier; it depresses me...)
Click to expand...
I am really a defeated person in life......I am weak in maths..... despite efforts i failed eariler to achieve my desired goal....... wish i could be like you ...
 
defeated_soldier said:
I am really a defeated person in life......I am weak in maths.....
Click to expand...
C'mon Buddy; EVERYBODY has a strong point(s),
and it sure doesn't have to be maths :idea:
 
Denis said:
Cause C being the 90degree corner and c being the hypotenuse is STANDARD; the [student] was using B.
Click to expand...
I have a feeling that the student was using whatever it was that the exercise showed.

Just my opinion, of course; I could be wrong....

Eliz.
 
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