Question

ABCD is a rhombus. RABS is a straight line such
that RA = AB = BS. Prove that RD and SC when
produced meet at right angles. ​

Answer 1

Answer:

Answer(s)

∠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 because ABCD is a rhombus. 

So RA = DA (both equal to AB) 

and BC = BS (both equal to AB). 

Triangle RAD is an isosceles triangle with RA=DA, 

so base angle ARD measures (180° - m∠RAD)/2. 

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

so base angle BSC measures 

(180° - m∠CBS)/2. 

Thus, 

m∠ARD + m∠BSC 

= (180° - m∠RAD)/2 + (180° - m∠CBS)/2 

= [360° - (m∠RAD + m∠CBS)]/2 

= (360° - 180°)/2 [because ∠RAD and ∠CBS are supplementary] 

= 90° 

Let T be the point where RD and SC intersect. 

Then 

m∠SRT + m∠RST 

= m∠ARD + m∠BSC [same angles] 

= 90° 

and the third angle of triangle RST, at T, measures 

180° - 90° = 90° 

which is what we wanted to prove.

Answer 2

Step-by-step explanation:

♤ ∠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 because ABCD is a rhombus. 

So RA = DA (both equal to AB) 

and BC = BS (both equal to AB). 

Triangle RAD is an isosceles triangle with RA=DA, 

so base angle ARD measures (180° - m∠RAD)/2. 

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

so base angle BSC measures 

(180° - m∠CBS)/2. 

Thus, 

m∠ARD + m∠BSC 

= (180° - m∠RAD)/2 + (180° - m∠CBS)/2 

= [360° - (m∠RAD + m∠CBS)]/2 

= (360° - 180°)/2 [because ∠RAD and ∠CBS are supplementary] 

= 90° 

Let T be the point where RD and SC intersect. 

Then 

m∠SRT + m∠RST 

= m∠ARD + m∠BSC [same angles] 

= 90° 

and the third angle of triangle RST, at T, measures 

180° - 90° = 90°