Proposed Problem

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

See also:

Complete Problem 303

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Sunday, June 14, 2009

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

Labels:
45 degrees,
90,
angle bisector,
degree,
triangle

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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.

ReplyDeleteThus, 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

Let BC = a

ReplyDeleteCase 1) Triangle ABC is scelene.

angle BAD = 45

So,

(BD)*(BD) = c*c + x*x - (square root of 2)*c*x

angle CAD = 45

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.

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

ReplyDeleteDraw DE perpendicular to AC, => ADE has DE=AE

ReplyDelete=>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

[ABC]=[ACD]+[ADB]

ReplyDeletethen

bc=bxsin45+cxsin45

dividing both sides by bcxsin45 gives the solution.

.-.