# FOUR geometry prob..

• February 4th 2007, 04:16 PM
gamecube10074
FOUR geometry prob..
I really really really need prob 2 and 4 answered first, then 1 and then 3 :P. PLEASE STATE ALL STEPS ONE BY ONE, 1. reason reason reason 1. GIVEN or... vertical angles contrugent, like that.

problem 1: angle E of triangle ZEN is obtuse. The bisector of angle E intersects SEGMENT ZN at X. J and K lie on segment ZE and sgement NE with ZJ=ZX and NK=NX. Discover and prove something about quadrilateral ZNKJ

problem 2: given: Segment FD parallel segment AC, segment BC parallel segment AE, segment FB parallel segment EC
Prove: B, D, and f midpoints of AC, CE, and EA.
http://img143.imageshack.us/img143/3...ngleuv9.th.png[/QUOTE]

problem 3: lengths of the sides of triangle ABC are BC=12, CA= 13, AB= 14. If M is mp of CA and P is point where CA cuts by the bisector of angle B, find the length of MP.

problem 4: PRove Ceva's ThM: If P is any point inside triangle ABC, then AX over XB times BY over YC times CZ over ZA = 1.

(( THESE HINTS ARE NOT GIVEN!!! ONLY THE AUX LINES ARE GIVEN AND YOU MUST STATE THEM.hints: use the aux lines, they are parallel to segment CX... apply triangle proportionality thm to triangle ABM, show that triangle APN is ~ to traingle MPB, traignel BYM ~ triangle CYP and triangle CZP ~ triangle AZN))
http://img142.imageshack.us/img142/7...athmgk6.th.png
• February 5th 2007, 12:32 AM
AlvinCY
Question 2:

This can be done easily with some properties of parallel lines and transversals.

Looking at the diagram, let the ratio of ED : DC be p : (1-p)... p can be thought of as a "factor" of the length EC. and 1-p is the complement factor as ED + DC = EC.

By properties of parallel lines (AE//BD), AB : BC = p : (1-p)

Once again by the properties of parallel lines (BF//CE), AF : FE = p : (1-p)

BUT... by the same argument, using FD//AC, we know that AF : FE = CD : DE

Therefore (1-p) : p = p : (1-p), and this can only occur if the ratio 1-p = p that is p = 1/2 (this means that ED : DC = AB : BC = AF : FE = 1/2 : 1/2 = 1 : 1) meaning that B, D, F are indeed midpoints of the 3 corresponding sides.
• February 5th 2007, 01:53 AM
gamecube10074
*easy x.x*
Thanks man.....but i don't think we know the way of solving in which you are using, which would be a total dead give away x.x
I STILL WOULD LIKE ANSWERS SEEING THEY MIGHT BE ON THE TEST, simplist way x,x
• February 5th 2007, 02:20 AM
AlvinCY
Hey hey!

After dinner, my brain works a little better and Figured out Question 3! Question 1 and 4 still stumps me.

So... hoping you know the sine rule... which states that if in a triangle, the angles are A, B and C, and a, b, c being the corresponding OPPOSITE side of A, B and C respectively, then:

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$

We will let angle ABP and angle CBP have a size of $T^\circ$

So... looking at triangle ABP, we know that:

$\frac{AP}{sin T^\circ} = \frac{BP}{sin A}$

We alos know from looking at triangle CBP that

$\frac{CP}{sin T^\circ} = \frac{BP}{sin C}$

We can see that these two equations give:

$sin T^\circ \times BP = PC \times sin C = AP \times sin A$

This gives me,

$\frac{AP}{PC} = \frac{sin C}{sin A}$

So if we work out what $\frac{sin C}{sin A}$ is, we can work out the ratio AP:PC

Using Triangle ABC and sine rule once again, we see that

$\frac{14}{sin C} = \frac{12}{sin A}$

Therefore, $\frac{sin C}{sin A} = \frac{14}{12}$

So, AP:PC = 14:12 = 7:6, so AP = 7cm, PC = 6cm, doing a simple subtraction: PM = 7cm - 6.5cm = 0.5cm!!!
• February 5th 2007, 03:08 AM
topsquark
Quote:

Originally Posted by gamecube10074
THANKS!! +REP TO ANYONE WHO DOES ANY OF THESE, post here ASAP!!!!!

I find this comment to be offensive. Anyone who helps you deserves "rep" or "thanks." If you choose to do so, then fine. If you choose not to do so, then fine. The choice is yours. But please don't try to make it some sort of payment. I consider that to be an abuse of the system.

-Dan
• February 5th 2007, 10:15 AM
gamecube10074
sorry top, i erased that part, i didn't know that was allowed...
alven....
The book says the answer is 0.5, but i didn't know how to get it :). ((i asked a friend how to do it, :) )
• February 5th 2007, 12:19 PM
Soroban
Hello, gamecube10074!

Here's #3 . . .

Quote:

3) Lengths of the sides of triangle $ABC$ are: $BC=12,\:CA= 13,\:AB = 14$
If $M$ is the midppoint of $CA$ and $P$ is point where $CA$ is cut by the bisector of $\angle B$,
find the length of MP.

Code:

                A                 o               *  *               *    * 13-x             *        *             *          *         14 *              *  P           *                o         *              *    *  x         *        *            *       *    *                    *     B o  *  *  *  *  *  *  *  *  *  o C                     12

$BP$ is the bisector of $\angle B:\;\angle ABP = \angle PBC$

Let $x = PC$, then $AP = 13-x$

Theorem: An angle bisector divides the opposite side into two segments
. . . . . . . proportional to the two adjacent sides.

So we have: . $\frac{x}{13-x} = \frac{12}{14}\quad\Rightarrow\quad x \,=\,PC \,= \,6$

Since $M$ is the midpoint of $CA,\;MC = 6.5$

Therefore: . $MP \:=\:MC - PC \:=\:6.5 - 6 \:=\:\boxed{0.5}$

• February 5th 2007, 03:34 PM
AlvinCY
Quote:

Originally Posted by AlvinCY

Therefore, $\frac{sin C}{sin A} = \frac{14}{12}$

So, AP:PC = 14:12 = 7:6, so AP = 7cm, PC = 6cm, doing a simple subtraction: PM = 7cm - 6.5cm = 0.5cm!!!

If you read it carefully, I actually gave you the answer as 0.5cm.

Cheers.
• February 5th 2007, 11:17 PM
AlvinCY
By the way, often wikipedia has some good stuff:

Ceva's theorem - Wikipedia, the free encyclopedia

LoL

I'm still stumped on Question 1, although, using a drawing program, I can see that the line segment JK is always parallel to ZN... but proving it is the difficult part.

So at this stage, I can say that it's a trapezium. that's it.