Similar Triangles and Area
Date: 11/12/98 at 04:32:11 From: Anna Subject: Geometry Hi, 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 http://mathforum.org/dr.math/
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 triangle: A /|\ / | \ / | \ / | \ / | \ D/______P___|_____\E /| /|\ 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 http://mathforum.org/dr.math/
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