Question

The side AB of a parallelogram ABCD is produced to X and the bisector of ∠CBX meets DA produced and DC produced at Y and Z respectively, prove that : DY = DZ = AB + BC.

Answer 1

Given:

ABCD is a parallelogram.

AB is produced to X, so X lies on the line AB beyond B.

The bisector of ∠CBX meets the line DA produced at Y and the line DC produced at Z.

To Prove:

DY = DZ = AB + BC.

Proof (Step-wise):

1. Put coordinates:

Let A = (0, 0). 

Put AB along the x-axis so B = (a, 0) where a = AB.

Let the vector AD = (p, q).

Then D = (p, q) and C = B + AD = (a + p, q).

Denote b = |BC|

= |AD|

= `sqrt(p^2 + q^2)`

In a parallelogram AD = BC and AB = DC.

2. Parametrize BX:

Since X is on the ray through B in the direction of BA the positive x-direction, the unit vector along BX is (1, 0).

The unit vector along BC is `(p/b, q/b)`.

3. By the internal angle-bisector property, the bisector direction from B is along the vector v 

= `(1, 0) + (p/b, q/b)`

= `(1 + p/b, q/b)` 

So, the bisector line ℓ through B has parametric equation ℓ:

`(x, y) = (a, 0) + s(1 + p/b, q/b), s ∈ R`

4. Intersection with DA produced (point Y).

The line DA produced is the line through A in the direction A→D = (p, q).

Points on DA produced can be written as A + t(–p, –q) = (–tp, –tq) for t ∈ R (t > 0 gives extension beyond A).

Solve `(a, 0) + s(1 + p/b, q/b) = (-t p, -t q)`.

From the y-coordinate:

`s(q/b) = -tq`

⇒ s = –tb 

From the x-coordinate:

`a + s(1 + p/b) = -t p`

Substitute s = –tb:

`a - tb(1 + p/b) = -tp`

⇒ a – t(b + p) = –tp 

⇒ a – tb – tp = –tp 

⇒ a – tb = 0 

⇒ `t = a/b`

Thus, Y = (–tp, –tq) with `t = a/b`.

Therefore, DY = Distance between D = (p, q)

And `Y = (-(a/b)p, -(a/b)q):` 

DY = |D – Y|

= `|(p, q) - (-(a/b)p, -(a/b)q)|`

= `|(1 + a/b)(p, q)|`

= `(1 + a/b) xx b`

= b + a

= AB + BC

5. Intersection with DC produced (point Z).

The line DC produced is the line through D in the x-direction:

D + u(a, 0) = (p + au, q) for u ∈ R.

Solve `(a, 0) + s(1 + p/b, q/b) = (p + au, q)`.

From the y-coordinate:

`s(q/b) = q`

⇒ s = b 

From the x-coordinate:

`a + b(1 + p/b) = p + au`

⇒ a + b + p = p + au 

⇒ a + b = au 

⇒ `u = 1 + b/a`

Thus, Z = (p + au, q)

= `(p + a(1 + b/a), q)`

And the distance DZ = |au| 

= `a(1 + b/a)` 

= a + b

= AB + BC

6. Hence, DY = DZ = a + b = AB + BC.