# Thread: Geometry in a parallelogram

1. ## Geometry in a parallelogram

Can someone please explain to me in an easy way to understand how i can do these proof questions, i really hate geometry.

2. Originally Posted by Misa-Campo
Can someone please explain to me in an easy way to understand how i can do these proof questions, i really hate geometry.

To show AXM and BXC are similar we need only show that they have two equal (congruent) angles.

1. Angles MXA and CXB are congriuent as they are vertical angles.

2. Angles AMX and CBX are congruent as they are alternate interior angles for the line BM between the parallel lines AD and BC.

Hence triangles AXM and BXC have two congruent angles (and so the third angles in each triangle are also congruent to one another as the angle sum in any triangle is 180 degrees) and so are similar.

RonL

3. Originally Posted by Misa-Campo
Can someone please explain to me in an easy way to understand how i can do these proof questions, i really hate geometry.

RonL

4. Originally Posted by Misa-Campo
Can someone please explain to me in an easy way to understand how i can do these proof questions, i really hate geometry.

(iii) Apply Pythagoras's theorem to triangle ACB and then use (ii).

AB^2+BC^2=AC^2

but BC=AB/2, so:

AB^2(1+1/4)=AC^2

but AC=(3/2)CX, so:

AB^2(1+1/4)=(9/4) CX^2

and the rest is just algebra.

RonL

5. Originally Posted by CaptainBlack

RonL
Yea m is the midpoint AD sorry forgot to put that in

6. Hello,

For ii), use i)

Since AXM and BXC are similar, the ratio of the sides is the same :

AX/XC=AM/BC