Question

ABCD is a parallelogram. BT bisects angle ABC and meets AD at T. A straight line through C and parallel to BT meets AB produced at P and AD produced at R. Consider two statements (1) triangle RAP is isosceles, (2) Sum of two equal sides of triangle RAP is equal to the perimeter of the parallelogram ABCD. Which of the statement is correct (a) Only 1 (b) Only 2 (c) Both 1 and 2 (d) None of 1 and 2

Answer 1

Option C:

Both (1) and (2)

Solution:

The image of the diagram is attached below.

$D C \parallel A B \text { or } D C \parallel A P$

$\therefore D C=\frac{1}{2} A P$ by mid-point theorem,

$\begin{array}{l}{D C=A B=B P} \\{C B \parallel A D \text { or } C B \parallel R A}\end{array}$

$\therefore C B=\frac{1}{2} R A$  by mid-point theorem,

$C B=D A=D R$

Let ABCD be a square  (Since square is also a parallelogram)

$\therefore D C=C B=B A=D A$

Here, $D C=C B$ and $D C=\frac{1}{2} A P \ldots(1)$

$C B=\frac{1}{2} A R \ldots(2)$

Compare equation (1) and equation (2),

$\Rightarrow \frac{1}{2} A P=\frac{1}{2} A R$

Multiply by 2 on both side of the equation.

$\Rightarrow A P=A R$

∴Δ RAP is an isosceles triangle.

To show that AP + AR = Perimeter of the parallelogram

AB + BC = AP and AD + DC = AR

Perimeter of ABCD = AB + BC + CD + DA

                                = AP + AR

Perimeter of ABCD = AP + AR

Sum of two equal sides of triangle RAP is equal to the perimeter of the parallelogram ABCD.

Option C is the correct answer.

Both (1) and (2) are correct.