## Tuesday, August 10, 2010

### Problem 497: Triangle, Two Squares, Centers, Midpoint, 90 Degrees

Geometry Problem
Click the figure below to see the complete problem 497 about Triangle, Two Squares, Centers, Midpoint, 90 Degrees, Congruence.

Complete Problem 497

Level: High School, SAT Prep, College geometry

1. ON midline of ▲AGC
OM midlime of ▲ADC
AG = DC , AG perpendicular DC
=>
OM = ON, OM perpendicular to ON

2. I see AG=DC. I don't understand that AG perpensicular DC. Excuse me please explain.

3. 1. Add the other midpoints in for symmetry: X on AB and Y on BC. This forms
a parallelogram XBYO and 3 similar triangles ABC, COY and AOX.
2. tr MOX is congruent to tr NOY by SAS. (Note: they are inverted versions) OX = BY from the parallelogram. OY = BX for the same reason. NY = BY since BNY is a right isosceles. Similarly MX = BX. The middle angles (MXO and NYC are congruent from an angle chase: 90 + angle B in both cases.)
3. This gives you that OM = ON.
4.That also shows that angle XOM and YON are two of the three angles in the
congruent triangles. So XOM + YON = 180 - (OXM or OYN). Each of these angles is 90 + angle B. And from the parallegoram angle XOY is also angle B.
So putting those together angle MON = 180 - (90 + angle B) + angle B = 90.

4. ∆ABG≡∆DBC (AB=DB, BG=BC), ∡ABG = ∡CBD) => AG = DC => ON = OM
From congruence, ∡CDB = ∡AGB ( perp sides ) => AG ⊥ DC, => ON ⊥ OM

5. From my solution to problem 496, AG = CD and so OM = DC/2 =AG/2 = O using the mid point theorem on Tr.s ADC and AGC.

Also from my solution to Problem 496,CD is perpendicular to AG and so
< GAC + < ACD = 90.
Since < GAC = < NOC and < ACD = < MOA,
< NOC + < MOA = 90 and so < MON = 90

Sumith Peiris
Moratuwa
Sri Lanka