Question

14). In ΔPQR and ΔDEF, if PQ:XY=PR:ZX,

Then ΔPQR∼ΔDEF when In ΔPQR and ΔDEF, if

PQ:XY=PR:ZX, then ΔPQR ΔDEF when​

Answer 1

Answer:

Let AD,BE and CF meet at O. We take O as the origin.

Let the coordinates of points A,B and C be (x

1

,y

1

), (x

2

,y

3

) and (x

3

,y

3

), respectively.

And let D divide BC in the ratio k:1 i.e.,

DC

BD

=

1

k

Then, by intersection formula

D≡(

k+1

k(x

3

)+1(x

2

)

,

k+1

k(y

3

)+1(y

2

)

)

D≡(

k+1

kx

3

+x

2

,

k+1

ky

3

+y

2

)

Also A and O lies on line AD .

So, equation of the line AD is

y−0=(

x

1

−0

y

1

−0

)(x−0)

or

y=

x

1

y

1

x (i)

SinceDliesonAD,itwillsatisfyequation(i)$$

So,

k+1

ky

3

+y

2

=

x

1

y

1

(

k+1

kx

3

+x

2

)

kx

1

y

3

+x

1

y

2

=kx

3

y

1

+x

2

y

1

⇒k(x

1

y

3

−x

3

y

1

)=x

2

y

1

−x

1

y

2

⇒k=

DC

BD

=

x

1

y

3

−x

3

y

1

x

2

y

1

−x

1

y

2

(ii)

Similarly, we can prove for line CA and AB that,

EA

CE

=

x

2

y

1

−x

1

y

2

x

3

y

2

−x

2

y

3

(iii)

and

FB

AF

=

x

3

y

2

−x

2

y

3

x

1

y

3

−x

3

y

1

(iv)

From (i),(ii) and (iii), we get

DC

BD

×

EA

CE

×

FB

AF

=1

or

BD×CE×AF×=DC×EA×FB

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