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Similar Triangles and Area

Date: 11/12/98 at 04:32:11
From: Anna
Subject: Geometry


I'm having difficulty with this question: 

P is a point on the segment joining the respective midpoints, D and E 
of the sides AB and AC of a triangle ABC. Prove that the area of BPC is 
twice the area of ADE.

I've drawn a diagram but I'm not completely sure what to do. Some 
people are using the midpoint theorem, and I have been reading some 
notes we have on Euclidean Geometry in which there is a similar problem 
dealing with congruent triangles. 

Thanks for your time. I hope you can help me.

TimeStamp: 11/12/98 at 06:28:50
From: Doctor Mitteldorf
Subject: Re: Geometry

Dear Anna,

First, notice that ADE and ABC are similar triangles, because they have 
one angle in common and their sides are in the ratio 1:2. It follows 
that DE is parallel to BC, and half its length. So the base BC of BCP 
is half the base DE of DEA. It remains only to show that the heights of 
these two triangles are the same. Do this by dropping a perpendicular 
from A down to line DE. It will be perpendicular to BC as well. Use 
similar triangles to show that the perpendicular dropped from A is the 
same length as the perpendicular dropped from P to BC.

That's just an outline, but I trust you can fill in the details. Let me 
know if there are any sticky points.

- Doctor Mitteldorf, The Math Forum   

Date: 11/17/98 at 03:41:58
From: Anna
Subject: Could I have some more help please?

I tried to follow the hints on my diagram but I can't seem to see how 
it works. At first I thought that my points were labelled differently 
so I changed them and then saw that the point P doesn't fit the 
triangle properly, and the original components in the question are 
then incorrect.

Would it be possible to have a little more help please?

Date: 11/17/98 at 13:28:17
From: Doctor Rick
Subject: Re: Could I have some more help please?

Hi, Anna. Let's draw a figure so we know we are talking about the same

                        /  | \
                      /    |  \
                    /      |   \
                  /        |    \
              /|      /|\  F      \
            /  |   /   |   \       \
          /    |/      |      \     \
        /    / |       |        \    \
      /   /    |       |           \  \
   B           G       H                C

Have you learned the theorem that the line joining the midpoints of two 
sides of a triangle (DE) is parallel to the third side (BC)? I think 
you must have, and the other Doctor showed you how to prove it.

You can use similar triangles (ADE and ABC) and the fact that 
AD = (1/2) AB to prove that DE = (1/2) BC. This gives you the relation 
between the bases of the two triangles ADE and BPC.

In order to prove the relation between the areas, you also need to look 
at the altitudes of the two triangles. Draw the altitudes, AF and PH, 
and also draw DG perpendicular to BC. You can prove that triangles 
ADF and DBG are congruent, and from this prove that AF = DG. Because 
DE is parallel to BC, you know that altitude PH = DG.

I have given you all the pieces of the proof. Can you put them 
together now?

- Doctor Rick, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
Middle School Geometry
Middle School Triangles and Other Polygons

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