# In desperate need of help! 3 Problems

• Jun 18th 2007, 07:11 AM
arden
In desperate need of help! 3 Problems
My Discrete exam is coming up in two days and I just realized that I didn't copy down solutions for a really hard proofs test that I had. These are three questions that I had no idea how to do on my test. Please help!

1. Given: ABCD is a parallelogram with equilaterals ABF and ADE drawn on sides AB and AD, respectively. Prove that triangle FCE is equilateral.

2. P is any point on rectangle ABCD. Prove that PA^2 + PC^2 = PB^2 + PD^2

3. In triangle ABC, altitude BE is extended to G so that EG = the measure of altitude CF. A line through G and parallel to AC meets BA extended at H, as shown in the figure. Prove that AH = AC.

Diagrams are attached (in order)! Thanks alot!
• Jun 18th 2007, 10:13 AM
Krizalid
For the first one

$\displaystyle \triangle{AEF}\cong\triangle{DEC}\implies\overline {EF}=\overline{CE}$, but $\displaystyle \measuredangle~AEF=\measuredangle~CED\implies\meas uredangle~CEF=60^\circ\,\therefore\,\triangle{FCE} \text{ is equilateral}\,\blacksquare$
• Jun 18th 2007, 10:19 AM
Krizalid
For the second one

Draw a line through $\displaystyle P$ such that cuts $\displaystyle \overline{AB}$ and $\displaystyle \overline{CD}$ orthogonally.

Then play with Pythagorean Theorem.
• Jun 18th 2007, 10:50 AM
Krizalid
Finally, for the third one, I'll give you a

Hint

Note that $\displaystyle \triangle{ABE}\sim\triangle{HBG}$ and $\displaystyle \triangle{ACF}\sim\triangle{ABE}$

Make a fusion with this equalities to get what you want.

See ya :)