## Monday, May 16, 2011

### Geometry Problem 602: Triangle, Angles, 30 and 45 Degrees, Cevian, Congruence

Geometry Problem
Click the figure below to see the complete problem 602.

1. w.l.o.g. assume AB = DC = 2
Draw BN ⊥ AC
Easy to see BN = NC = 1,
Follows DN = DC – NC = 2 – 1 = 1
BN =DN = CN =1 implies
ΔDBC is right angled isosceles and BD = √2
Now apply Sine Rule to ΔADB:
Sin ∠ADB: sin 30° = AB: BD = 2: √2
√2 Sin (30° + x) = 2 sin 30°= 1
Sin (30° + x) = 1/√2 = sin 45°
x = 15°

2. Rather than an elegant plane geometry solution, if we can use Sine Rule then no construction is really necessary for BC/AB=sin(30)/sin(45)= 1/√2 & BC/CD=sin(45)/sin(105-x). In other words,sin(105-x)=1 since AB=DC or x=15 deg. Isn't that so?
Vihaan

3. Referring to my earlier solution:

ΔDBC is right angled isosceles implies
30° + x = ∠ADB = ∠ABD = 45°, x = 15°
(No need to apply Sine Rule)

Pravin

4. There's no need to assume AB = CD = 2, we can just put AB = CD = 2a and that makes the solution much valid.

5. La idea es resolverlo con elementos de euclides y no por trigonometria con teoremas del seno y del coseno...

6. CE=BE=1/2AB where BE is the altitude of Tr ABC. But DE= CD-CE=CE. So <DBE =45 and hence x=15

Sumith Peiris
Moratuwa
Sri Lanka