Sum to infinity question (G.P.)

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The discussion revolves around finding the sum to infinity of a geometric series formed by the lengths CA, AD, DE, and subsequent perpendiculars in triangle ABC. The triangle is identified as a right triangle, allowing the use of Pythagorean theorem and properties of similar triangles to derive the lengths of the perpendiculars. Initial assumptions about the perpendicular bisecting the base were corrected, emphasizing the need for accurate geometric reasoning. Participants suggest sketching the triangle for better visualization and applying trigonometric relationships to find the lengths. Ultimately, the problem is clarified, leading to a realization that the solution is more straightforward than initially perceived.
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Homework Statement



ΔABC has AB= 8 in, BC= 10in , CA= 6in . AD is the perpendicular from A to BC. DE the perpendicular from D to AB, EF the perpendicular from E to BD and so on. Show that CA + AD + DE+... is a geometric series and find it's sum to infinity.

Homework Equations





The Attempt at a Solution



Umm I assumed that the perpendicular would halve the base and I used Pythagoras' theorem to find the sides I would need. Well this turned out to be incorrect. I'm not sure how to find the right lengths of the sides, my geometry is weak. Could someone guide me and also give me a site or something that has a good geometric problems.
 
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lionely said:

Homework Statement



ΔABC has AB= 8 in, BC= 10in , CA= 6in . AD is the perpendicular from A to BC. DE the perpendicular from D to AB, EF the perpendicular from E to BD and so on. Show that CA + AD + DE+... is a geometric series and find it's sum to infinity.

Homework Equations


The Attempt at a Solution



Umm I assumed that the perpendicular would halve the base and I used Pythagoras' theorem to find the sides I would need. Well this turned out to be incorrect. I'm not sure how to find the right lengths of the sides, my geometry is weak. Could someone guide me and also give me a site or something that has a good geometric problems.

The perpendicular does not halve the "base". Your assumption is incorrect.

ΔABC is a right triangle. Can you show that?

ΔABD and ΔACD are also right triangles, by assumption. There are two ways to find the length of AD; (1) right angle trig along with the given lengths for the original triangle or (2) facts regarding similar triangles and the preservation of ratios of corresponding side lengths.

Finding the length of each subsequent perpendicular is done in a similar (no pun intended) way.

If you haven't already done so, you should definitely sketch a decent picture of what you are working with.
 
Now that you mention it , Triangle ABC would be right angled, cause 6^2 + 8^2 = 10^2

So the 1st perpendicular dropped makes Two isosceles triangles appear right? The perpendicular bisects the 90 at A?
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Look at the attached figure

Apply Pythagoras theorem to ΔADC and Δ ABD .You will get two equations with two unknowns (p and x) .Using them find the length of the perpendicular AD (i.e p).

Similarly consider ΔBDE and ΔDEA .Again apply Pythagoras theorem .Find the length of the perpendicular DE .

Do you see a pattern in the lengths CA ,AD, DE ?

What do you get ?
 

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Thanks so much, I realized that this was pretty easy, I guess I was being too lazy with the thinking..
 
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