What is the shortest way to solve this geometry problem?

Question

PQR is a triangle right angled at P and M is a point on QR such that PM┴QR. Show that PM² = QM * MR

I got the answer, but it took soooo many steps! No using similarity. We must only use the Pythagoras theorem. My teacher says we can do it in less than 6 steps! I used a whopping 13 steps to get it! Help. Is there a shorter way?

"NO USING SIMILARITY"
I wonder how many of you actually read the details properly beyond the query.

Prashant gets Best Answer. Drat, I must wait 4 hours first.

By the way, congrats on your Top Contributor badge, Prashant. I just noticed it, been busy lately. I knew it was coming from the moment you answered my mirror question in Physics. You were the only one to answer it, remember?

Answer 1

PR^2 = PM^2 + MR^2
PQ^2 = PM^2 + QM^2

Add :
QR^2 = 2PM^2 + MR^2 + QM^2
(QM + MR)^2 = 2PM^2 + MR^2 + QM^2
QM^2 + MR^2 + 2QM.MR = 2PM^2 + MR^2 + QM^2

Thus
PM^2 = QM.MR

Hope this helps.

[e-mail]

Answer 2

First, show that triangles RMP and PMQ are similar to RPQ. That is straightforward, since all three triangles have a right angle and RMP and PMQ both share an angle with RPQ.

Since both RMP and PMQ are similar to RPQ, it follows that they are similar to each other. Knowing that, we can set up the ratio MR / PM = PM / QM, and so PM² = QM * MR.

* * * * *

Prashant's use of the Pythagarean theorem also works -- actually, it's kind of nice how the terms cancel out to get the answer. However, I still think using similar triangles is easier -- you did ask for the fastest way, right. :)

Answer 3

this would not sound suited are you able to probable grant a image or something? If the two shorter factors of the triangle are 9 and 18, then that dictates the size of the hypotenuse via R = sqrt ( 9^2 + 18^2 ) = 20 for the longest facet. i've got on no account seen a ramp that wasn't a suited triangle.

Answer 4

From triangle QPR:
QR^2 = QP^2 + PR^2
from triangle PMR
PR^2 = PM^2 + MR^2
Substitute:
QR^2 = QP^2 + PM^2 + MR^2
0 = QP^2 + PM^2 +MR^2 -QR^2 **

from triangle QPM:
QP^2 = QM^2 + PM^2
0 = QM^2 + PM^2 - QP^2 **
Add the starred eqn's
2PM^2 + MR^2 - QR^2 + QM^2 = 0
2PM^2 = - MR^2 + QR^2 - QM^2
but QR = QM + MR
2PM^2 = (QM + MR)^2 -QM^2 - MR^2
2PM^2 = 2* QM *MR
PM^2 = QM * MR

Answer 5

in PQM triangle:
P1 angle is front of QM and equal with R angle in PQR triangle:
P1+Q=90
R+Q=90
then p1=R
two triangle PQR and PQM are like then we can write:
PM/QM=MR/PM
then:
PM^2=QM*MR