Triangles doubt.?

Question

Triangle ABC is right-angled at B. D and N and points on AC and AB respectively such that DN is parallel to BC. Draw DM perpendicular to BC. Prove that:

DM² = (DN)*(MC)
DN² = (AN)*(DM)

Bracha, the points are fixed. D is on AC, N is on AB and M on BC. Read the question carefully.

CPUcate, you must have me wrong, but I believe I said D and N are points on AC and AB, not their midpoints. Same for M.

Joining midpoints is not the only way to get parallel lines in a right triangle! Why do you keep misunderstanding? Or did you not read Additional Details?

D and N can be anywhere on AC and AB respectively. You can still get parallel lines DN and BC. And naturally, it follows that M is not the midpoint of BC.

You guys are dealing with a former Top Contributor in Mathematics here.

OR if you need the source of this:
This is from Exercise 6.6, question 2 of the NCERT maths textbook for Class X (NEW). You can see the Online textbook here:
http://www.ncert.nic.in

Answer 1

it is only true when point D is in the midpoint of AC, and the triangle is an isosceles triangle..

let
AB = 2
BC = 2
D, N, and M are in the mid points of the line segments in which they lie in.. ~DN = 1 and ~DM = 1

then:
DM^2 = DN*MC
(1)^2 = (1)*(1)
1 = 1

DN^2 = AN*DM
(1)^2 = (1)*(1)
1 = 1

but it is not true on other triangles


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aditional details
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who cares if your FORMER top contributor.. i quote "former"... hehe..
have you not tried these formulas yet? if not i could prove to you that it those two formulas are wrong...

ok. lets start with an isoselis right triangle, with D not on the midpoint...

let
AB = 3
BC = 3
DM = 2
DN = 1

DM^2 = DN*MC
(2)^2 = 1*2
4 = 2
wrong 4 not equal to 2

DN^2 = AN*DM
(1)^2 = 1*2
1 = 2
again wrong 1 not equal to 2

ok.. next a acute right triangle still D not on the mid point..
let
AB = 4
BC = 3
AC = 5
DN = 2
DM = 8/3
(if you doubt any of these use distance formula and slope point formula)

DM^2 = DN*MC
(8/3)^2 = 2*1
64/9 = 2 or
7 and 1/19 = 2 still way too far and is wrong!

DN^2 = AN*DM
(2)^2 = (4/3)*(8/3)
4 = 32/9 or
4 = 3 and 5/9 still way too far and still wrong..


we understand your problem, and has proved it wrong. (maybe you got the formulas wrong) if you believe otherwise, try going to paper and pencil and try drawing the triangle you are saying... and if you found out that there are other triangles, beside for an isoseles right triangle with D on the midpoint, that can prove your triangle right pls email it to me.. and i will apologize.. coz i still think, its a hoax..

good luck!

Answer 2

No its not
let AB = 16 . . . BC = 8 . . . . . given
AN = 8 . . .. NB = 8 . . . . DM = 8
DN = 4 . . . . BM = 4 . . . . . MC = 4

DM² = (DN)*(MC) . . . 8² not equal to 4 (4)

DN² = (AN)*(DM) . . . 4² not equal to 8 (8)

Answer 3

I think we cannot prove it. Because the points D, N, and M are not fixed. Did I miss something?