Question

ABCD is a rhombus RABS is a straight line such that RA=AB=BS prove that RS and SC when produced meet at right angle

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Proved below.

Step-by-step explanation:

Given:

Here ∠DAB and ∠ABC are supplementary (adjacent angles in a rhombus).  

∠RAD and ∠DAB are supplementary (linear pair).  

∠ABC and ∠CBS are supplementary (linear pair).  

So ∠RAD and ∠ABC are congruent (supplementary to the same angle),  

and ∠RAD and ∠CBS are supplementary because ∠CBS is supplementary to an angle congruent to ∠RAD.  

DA = AB = BC = CD because ABCD is a rhombus.           [given]

So RA = DA (both equal to AB)  

and BC = BS (both equal to AB).  

Δ RAD is an isosceles triangle with RA = DA,  

so base angle ARD measures  $\frac{(180\degree - m\angle RAD)}{2}$.                                      [1]

Similarly, Δ CBS is an isosceles triangle with BC = BS  

so base angle BSC measures  $\frac{(180\degree - m\angle CBS)}{2}$.                                       [2]

Thus,  

m∠ARD + m∠BSC  

=  $\frac{(180\degree - m\angle RAD)}{2} + \frac{(180\degree - m\angle CBS)}{2}$             [ from 1 and 2 ]

= $\frac{360 - (m\angle RAD + m\angle CBS)}{2}$

= $\frac{360-180}{2}$   [because ∠RAD and ∠CBS are supplementary]  

= $\frac{180}{2}$

m∠ARD + m∠BSC = 90°                [3]

Let P be the point where RD and SC intersect.

Then  

m∠SRP + m∠RSP

= m∠ARD + m∠BSC            [same angles]  

= 90°                  [ from 3 ]

and the third angle of triangle RSP, at P, measures  

180° - 90° = 90°  

which is what we wanted to prove.  

Hence RS and SC when produced meet at right angle.