Parallel line in triangle.

[Helenam]

I am doing some extra work over the summer break but l am stuck with this question.

In the triangle ABC, two sides have the lengths: AB = 10, AC = 8. The bisector of the angle BAC intersects BC at point E. A line is drawn through point E parallel to side AB, intersecting side AC at point D. Calculate DE.
I sketched a right triangle ( it doesn’t say whether it is or isn’t) as shown in the attachment. Can anyone suggest where I go from here please.

 

[Dr.Peterson]

I am doing some extra work over the summer break but l am stuck with this question.

In the triangle ABC, two sides have the lengths: AB = 10, AC = 8. The bisector of the angle BAC intersects BC at point E. A line is drawn through point E parallel to side AB, intersecting side AC at point D. Calculate DE.
I sketched a right triangle ( it doesn’t say whether it is or isn’t) as shown in the attachment. Can anyone suggest where I go from here please.

You can't assume it's a right triangle; it turns out that the angle doesn't affect the answer, which is why they don't need to give you an angle.

On the other hand, given what I just said, you should get the correct answer if you do assume it's a right triangle! You'll learn more if you don't, though.

​

Have you learned any theorems involving angle bisectors in a triangle? If so, that is what you should use.

If not, you might know enough to the the method by which the theorem I have in mind is proved. Think about the ratio of areas of triangles AEB and AEC, which have a common base AE. Then think about segments EB and EC.

It may be helpful if you can tell us what you've learned in geometry.

 

[Helenam]

I redrew the triangle and used the angle bisector theorem which I have just found online. The answer I got is 40/9. I think this is correct? Thanks for your hint.

 

[Dr.Peterson]

I redrew the triangle and used the angle bisector theorem which I have just found online. The answer I got is 40/9. I think this is correct? Thanks for your hint.

Yes, that's right. You clearly had no trouble doing the additional steps (similar triangles and ratios) after applying that theorem!

 

[Helenam]

Yes, that's right. You clearly had no trouble doing the additional steps (similar triangles and ratios) after applying that theorem!

No I had no problems but I first did it with my right angled triangle as it was easier for me to visualise the similar triangles and ratios. Thanks again for your help.

 

[Pramod Kumar Tandon]

Since AE bisects angle A, Hence Angle BAE = A/2 = Alternate angle AED
Hence in triangle AED : Angle AED = A/2 = Angle DAE
Therefore, Side DE = AD
Now use Internal angle bi-sector theorem to get AD = DE = 40/9 and that's your answer.

 

[Helenam]

which angle are you bisecting when you say the internal angle bisector theorem. Do you mean 8/18 *10

 

[Dr.Peterson]

which angle are you bisecting when you say the internal angle bisector theorem. Do you mean 8/18 *10

Obviously it is angle BAC that is bisected. Pramod certainly didn't explain that part at all, which is perhaps the most important part. But, yes, one step in his method is to multiply 10 by 8/18.

There are other ways to get the answer besides the part Pramod showed. I myself multiplied 8 by 10/18, using similar triangles to get AD more directly, without using DE.

If you don't understand the proof, please show your work.

 

[Helenam]

I understood what you explained before but I assumed Pramod was introducing something new, hence the question about the angle. I bisected angle BAC as the original question stated and arrived at the answer.

 

[Dr.Peterson]

I understood what you explained before but I assumed Pramod was introducing something new, hence the question about the angle. I bisected angle BAC as the original question stated and arrived at the answer.

I have no idea why he bothered to say anything. All he did was to show that AED is isosceles, just as you did in your original picture! Then he used that in the solution, without showing the details of how. So he contributed nothing new.