Question

prove

Answer 1

why is your name highschoolmom?

Answer 2

this confused me because of the 3 right angle symbols in the pictures

Answer 3

There are several right triangles in the diagram. What have you studied lately in Geometry? I am thinking this problem has to do with similar triangles.

Answer 4

It has to do with similar triangles, but the question is, "What's the proper approach for the proof?".

Answer 5

similarity in right triangles

Answer 6

We know that much already of course.

Answer 7

@Directrix asked what ive been studying

Answer 8

http://www.coursesmart.com/SR/5548832/9781256493303/460?__hdv=6.8

that is my book

Answer 9

anyone have an idea

Answer 10

well lets here it :)

The proof is actually really short

Answer 11

*hear

Answer 12

i have no idea @Zarkon

Answer 13

use the two largest right triangles and similar triangles ... it will work out :)

Answer 14

how

Answer 15

im soo lost

Answer 16

concentrate on just this figure

Answer 17

Triangle ABC is similar to triangle DEC (both are right triangles, with = angles)

you can set up a ratio of corresponding legs

for example, \(\frac{\text{hypotenuse}}{\text{long leg} } = \frac{\text{hypotenuse}}{\text{long leg} }\)

Answer 18

can you do that ?

Answer 19

let me try

Answer 20

yes. now cross multiply to get

\[ AC \cdot EC = BC \cdot DC \]
now we do a trick. AC is the same as AD + DC and BC is the same as BE+EC
replace AC and BC in the equation:
\[ (AD + DC ) EC = (BE+EC) DC \]
now distribute both sides. can you do that ?

Answer 21

you would get
\[ AD \cdot EC + DC \cdot EC = BE \cdot DC + DC \cdot EC \]

subtract \( DC \cdot EC\) from both sides
\[ AD \cdot EC= BE \cdot DC \]
can you finish ?

Answer 22

ADEC=BEDC

Answer 23

yes, now divide both sides by DC and by EC to get what you are trying to prove

Answer 24

^^