## Friday, September 10, 2010

### Problem 519: Triangle, Squares, Altitude, Perpendicular, Rectangle, Area

Geometry Problem
Click the figure below to see the complete problem 519 about Triangle, Squares, Altitude, Perpendicular, Rectangle, Area. Complete Problem 519

Level: High School, SAT Prep, College geometry

1. ▲BPH ~ ▲ABH => PB/BH = BH/AB
▲BMH ~ ▲BHC => BM/BH = BH/BC
=> PB∙AB = BM∙BC

2. In right triangle ABH with altitude HP we have BH^2=BP.BA=S1 ( relations in right triangles)
Similarly in right triangle BHC we have BH^2=BM.BC= S2
So S1=S2

Peter Tran

3. Triangle ABH and BHP are similar --->
AB:BH=BH:BP ---> S1=ABxBP=BH^2

Triangle BCH and BHM are similar --->
BC:BH=BH:BM ---> S2=BCxBM=BH^2 ---> S1=S2

4. a trigonometric solution
BH=csinA=asinC
BM=BHsinC
BP=BHsinA
S1=c.BP=acsinAsinC
S2=a.BM=acsinAsinC
.-.

5. Angles BPM = BHM = C;
Triangles BPM, BCA are ///.
BP:BC = BM:BA,
BP.BA = BM.BC,
BP.BD = BM.BG
S1 = S2

6. Problem 519
Is BH^2=BP.BA=BP.BD=S_1, BH^2=BM.BC=BM.BG=S_2,Therefore S_1=S_2.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE