Math, asked by rohanspatil, 5 months ago

plz solve this question...​

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Answers

Answered by Yoursenorita
3

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.

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Answered by ItzStarling
1

Answer:

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