There is a rather hard Geometry problem, can you solve it?

Question

There is a group of Geometry students I am working with, and there is a homework packet we have to finish. There was a question we came across and we are unsure of how to solve it.

Prove: The sum of the lengths of the perpendicular segments drawn from any point in the base of an isosceles triangle to the legs is equal to the length of the altitude drawn to one of the legs.

We have to find the picture/figure for it first and then reason it out. Can anyone help us?

Answer 1

I drew a picture in MS paint which I can forward to you if you email me at a_math_guy at yahoo. The picture isn't that great. (I don't see attach as an option here.) Basically, imagine two long sides of an isosceles triangle connected at the bottom by a short base. Pick any point on the base (bottom) and draw perpendiculars to the two legs above it. This forms a sort of L shape with the joint of the L lying on the base. Add these two lengths. (That is one value.) The other value is the altitude that crosses (say) from the lower left hand point to the leg on the right. While I was drawing it in Paint I guessed at a possible proof (a first try), but your post indicates that you want to work on it some. I could give a hint for what I am thinking: the diagonals of a (symmetric? regular?) trapezoid are equal. Anyway, email me and I can forward what I drew in MSPaint as an attachment.

You could also try to solve it analytically: say the vertices for the traingle are (0,0), (2,0) and (1,y) and the point on the base is (x,0) with x between 0 and 2. Then solve all those distances using trigonometry and algebra.

I tried to load the file into my profile but I can't seem to do it without yahoo 360 and I just hate that so...............

Answer 2

Prove: The sum of the lengths of the perpendicular segments drawn from any point in the base of an isosceles triangle to the legs is equal to the length of the altitude drawn to one of the legs.

Diagram: http://i130.photobucket.com/albums/p248/WalC_photos/IsoscTriangle.jpg

Data: ΔABC is isosceles with AB = AC

Constructions:

Choose D on BC

Draw perpendiculars from D onto AB (meeting AB at E) and onto AC (meeting AC at F). and from B onto AC (meeting AC at F)

Join AD

Proof:

Area ΔABD = ½ AB * DE
Area ΔACD = ½ AC * DF

Also Area ΔABC = ½ AC * BG

But Area ΔABC = Area ΔABD + Area ΔACD
= ½ AB * DE + ½ AC * DF

Since ΔABC is isosceles, AB = AC (Data)

So ½ AB * DE + ½ AC * DF = ½ AC * DE + ½ AC * DF
= ½ AC (DE + DF)

Therefore ½ AC (DE + DF) = ½ AC * BG

So BG = DE + DF … QED

You can similarly show that this sum = length of altitude drawn from C onto AB ( as Area ΔABC = ½ AB * (altitude from C))

ie The sum of the lengths of the perpendicular segments drawn from any point in the base of an isosceles triangle to the legs is equal to the length of the altitude drawn to one of the legs

Thank you for this question ... I have never seen this before!!

Answer 3

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