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$\triangle ABC$ is right angled at $A$. $AB=20, CA= 80/3, BC=100/3$ units. $D$ is a point between $B$ and $C$ such that the $\triangle ADB$ and $\triangle ADC$ have equal perimeters. Determine the length of $AD$.

Inceptio
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Niharika
  • 1,051
  • What did you try? – rschwieb May 23 '13 at 10:17
  • I equated the perimeters to get the measures of line segments BD and CD. Then I took angle ABC as θ and evaluated the value of sin θ, which happens to be 4/5. From this I got the value of cos θ= 3/5. So since we know two sides and the value of cos of the included angle in triangle ABD, I used the law of cosines (c^2=a^2 + b^2+2abcosC) to get the value of c, which turns out to be 16/3.. I'm not quite sure if its a correct procedure though.. needed a verification on that..) – Niharika May 23 '13 at 10:40

4 Answers4

1

Hint: You can solve following these steps: enter image description here

newzad
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1

enter image description here

Add up the the sides of $\triangle ADB$ and $\triangle ADC$ and you get their perimeters and then equate them:

$x+\frac{100}{3}-y+\frac{80}{3}=x+y+20$

Solve for $y$ to get $y=20$

and then for $x$ using the cosine rule we get

$x=\sqrt{20^2+20^2-2(20)(20)\cos\left(\tan^{-1}\frac{80}{20\times3}\right)}$

$x=8\sqrt{5}$

Maazul
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0

Referring to zewzad's great sketch, I used Maazul's algebra to determine that

BD = 20

I then used coordinate vector geometry to obtain the coordinates of point D (defining the origin at A). This leads to coordinate values:

A = [ 0 , 0 ]

B = [ 0, 20 ]

C = [ 80/3, 0 ]

I used scalar multiplication and vector addition to obtain coordinates for D.

The vector that takes you from B --> C is < 80/3, -20 >. We know where point D lies fractionally going from B --> C. It's (20 / 100/3) or 0.6 of the full distance. Scalar-mult vector B-->C by 0.6 to get the vector B-->D:

B-->D = 0.6 * B-->C

B-->D = 0.6 * < 80/3, -20 >

  = < 16 , -12 >

by vector addition, point D has coordinates:

D = B + B-->D

= [ 0, 20 ] + < 16 , -12 >

= [ 16, 8 ]

Finally, distance AD can be computed as length of point-vector D = ( <16, 8> )

AD = sqrt (16^^2 + 8^^2 )

AD = sqrt (320)

AD = 17.8888 = 8*sqrt(5)

pbierre
  • 379
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Hint:

$AB+BD+AD=AD+CD+AC$

$AB+BD=CD+AC$

$AB+BD+DC=2CD+AC$

$20+\dfrac{100}{3}=2CD+\dfrac{80}{3}$

Solve for $CD$ .And then for $AD$.

Inceptio
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