plz solve this question...
Answers
GIVEN:
ABCD is a rhombus.RABS is a straight line such that RA=AB=BS.
TO PROVE:
Prove that RD and SC when produced meet at right angles.
SOLUTION:
- ∠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)
- ∠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:
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