Hello, waterqueen7!

Here's one approach . . .

Q 4. Rajan constructed rhombus ABCD.

He then drew a straight line RABS such that RA = AB = BS.

Prove that RD and SC when produced meet at a right angle. Code:

Q
*
* *
* *
D*-----------*C
*α/ /β*
* / / *
* / / *
*α /θ /θ β*
* - - - - *-----------* - - - - *
R A B S

We have: .$\displaystyle AB = BC = CD = DA = RA = BS$

Let $\displaystyle \theta = \angle DAB = \angle CBS$

.Then: .$\displaystyle \angle DAR = 180^o - \theta$

In isosceles $\displaystyle \Delta DAR$, let: .$\displaystyle \alpha = \angle DRA = \angle RDA$

Since the sum of its angles is 180°: .$\displaystyle (180^o-\theta) + 2\alpha \:=\:180^o\quad\Rightarrow\quad \alpha = \frac{\theta}{2}$ .**[1]**

In isosceles $\displaystyle \Delta CBS$, let: .$\displaystyle \beta = \angle CSB = \angle BCS$

Since the sum of its angles is 180°: .$\displaystyle \theta + 2\beta \:=\:180^o\quad\Rightarrow\quad \beta \:=\:90^o - \frac{\theta}{2}$ .**[2]**

In $\displaystyle \Delta QRS\!:\;\;\angle Q + \alpha + \beta \;=\;180^o\quad\Rightarrow\quad \angle Q \;=\;180^o - \alpha - \beta$

Substitute [1] and [2]: .$\displaystyle \angle Q \;=\;180^o - \frac{\theta}{2} - \left(90^o - \frac{\theta}{2}\right) \;=\;90^o$

Therefore: .$\displaystyle \angle Q$ is a right angle.