## Sunday, June 14, 2009

### Problem 303. Triangle, Angle bisector of 90 degrees

Proposed Problem
Click the figure below to see the complete problem 303 about Triangle, Angle bisector of 90 degrees.

Complete Problem 303
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

1. By sine rule in Tr. DAC, x/b = sin(C)/sin(B+45). Similarly, x/c = sin(B)/sin(C+45) = sin(B)/sin(B+45) since (B+45) and (C+45) are supplementary.
Thus, x/b + x/c = [sin(B) + sin(C)]/sin(B+45)=[(c/a+b/a)/sin(B+45)]= (b+c)/(asinB/V2 +acos(B)/V2) = V2(b+c)/(b+c) = V2 or 1/b + 1/c = V2/x
Ajit: ajitathle@gmail.com

2. Let BC = a
Case 1) Triangle ABC is scelene.
So,
(BD)*(BD) = c*c + x*x - (square root of 2)*c*x
So,
(CD)*(CD) = b*b + x*x - (square root of 2)*b*x

As AD is angle bisector of angle BAC,
BD/CD =c/b
So, [(BD)*(BD)]/[(CD)*(CD)] = [c*c] / [b*b]

By substituting value of (BD)*(BD) and (CD)*(CD) in this equation we get,
[c*c + x*x - (square root of 2)*c*x] /
[b*b + x*x - (square root of 2)*b*x]
= [c*c] / [b*b]
By cross multiplication we get,
b*b*c*c + b*b*x*x - (square root of 2)b*b*c*x
= b*b*c*c + c*c*x*x - (square root of 2)b*c*c*x

So,
b*b*x*x - (square root of 2)b*b*c*x
=c*c*x*x - (square root of 2)b*c*c*x

By dividing this equation by x,

b*b*x - (square root of 2)b*b*c
=c*c*x - (square root of 2)b*c*c

So,
b*b*x - c*c*x
= (square root of 2)*b*b*c
- (square root of 2)*b*c*c

So,
x*(b+c)(b-c) = (square root of 2)*b*c*(b-c)
As (b-c) is nonzero number (Reason : Case 1),
x*(b+c) = (square root of 2)*b*c
So,
x/(square root of 2) = (b*c)/ (b+c)
So,
(square root of 2)/x = (b+c)/(b*c)
= b/(b*c) + c/(b*c)
= (1/c) + (1/b)

Case 2) b = c or triangle ABC is isosceles triangle.

According to properties of an right isosceles triangle,
x = a/2

1/b + 1/c = 2/b

But b = a/(Square root of 2)
So,
1/b + 1/c = 2/[a/(Square root of 2)]
= 2 * (Square root of 2) / a
= (Square root of 2)/x

Hence the proof.

3. http://img379.imageshack.us/img379/3304/coz.gif

4. Draw DE perpendicular to AC, => ADE has DE=AE
=>x'2=DE'2+DE'2 or x'2=2DE'2
ABC and DEC are similar (AB//DE)
=>DE/c = b-DE/b => DE = cb / c+b

substitute DE => x'2 = 2 cb / c+b '2
give the result

P.S. '2 mean exponent 2

then
bc=bxsin45+cxsin45
dividing both sides by bcxsin45 gives the solution.
.-.

6. Complete the square APDQ, P on AB and Q on AC

S(ABC) = bc/2 = (c/2)x/V2 + (b/2)x/V2

So V2/x = 1/b + 1/c

Sumith Peiris
Moratuwa
Sri Lanka