Geometry Insight

Question

Hello,

I have a question for which I don't understand how the solution came about. In the above triangle, AC = 8, and BE = 6. AD and DE are the equal. Also CN and NB are equal i.e. N is the mid point of BC. The question is what is DN?

The solution starts by stating, create a line NF, where N is in between E and C and parallel to BE.

From here Triangle ECB is similar to FCN with a ratio of 2:1 so FN is 3. Then DF is equal to DE + EF. DE=1/2AE and EF=1/2EC So DE+EF = 1/2(AE+EC) = 1/2(AC) = 4. FN is 1/2 BE = 3. So DFN is 345 right triangle. So DN is 5.

I have a 2 part question. First, it didn't occur to me to draw that parallel line but what in the question would suggest that this is a good idea? I tried putting the triangle in a co-ordinate system but didn't get anywhere. I don't understand the thinking to get to the point where that this seems like a good idea.

The second, is there a different solution that would seem more straightforward?

Thanks,

Answer 1

For a different straightforward solution, set up the coordinate geometry with $ E = (0,0) $ (why is this a good choice?) $A = (0, 2b), B= (6, 0), C = (0, 2b-8)$.

Show that $D = (0, b), N= (3, b-4)$ and hence $|DN| = 5$.

Answer 2

Here is a variation of the proof, using the midline theorem extensively. Let M be a midpoint if AB. Then due to the midline theorem:

• MN is parallel to AC,
• MN is AC/$2=8/2=4$,
• DM is parallel to EB,
• DM is EB/$2=6/2=3$.

Since MN$\parallel$AC$\bot$EB$\parallel$DM, we have MN$\bot$DM. Now, from the right-angle triangle MND we get DN$=5$.