show that the angle bisector of co interior angle are perpendicular
Answers
Explanation:
In the given figure AB || CD. Transversal PQ
Given: Ray RQ bisects angle R and PS bisects angle S.
angle S and angle R are co interior angles.
To prove: RN perpendicular to SM.
and MS perpendicular to RN.
therefore angle N and angle M = 90°
Proof:
we know angle R and angle S are co interior angles.
So, angle R + angle S = 180°
1/2 angle R + 1/2 angle S = 180° x 1/2
angle MRS + angle MSR= 90° ------(1)
( reason: given PS bisects angle S and RQ bisects R, angle ARN = angle MRS and angle MSR = angle MSC.)
Now,
By A.S.P of triangle RMS,
angle MSR + angle MRS + angle RMS = 180°-----(2)
substituting eq.(1) in eq.(2),
(MSR + MRS )+ RMS= 180°
90° + angle RMS= 180°
angle RMS = 180° - 90°
angle RMS = 90°
similarly, by A.S.P of triangle RNS,
angle RNS + angle NRS + angle RSN = 180°
RNS + (NRS + RSN)= 180°
RNS + 90°=180°
RNS= 180° - 90°
RNS = 90°
thereby,
angle RMS = RNS = 90°
So, RM perpendicular to SN and SN perpendicular to RM.
HENCE PROVED.
Answer:
Explanation:
In the given figure AB || CD. Transversal PQ
Given: Ray RQ bisects angle R and PS bisects angle S.
angle S and angle R are co interior angles.
To prove: RN perpendicular to SM.
and MS perpendicular to RN.
therefore angle N and angle M = 90°
Proof:
we know angle R and angle S are co interior angles.
So, angle R + angle S = 180°
1/2 angle R + 1/2 angle S = 180° x 1/2
angle MRS + angle MSR= 90° ------(1)
( reason: given PS bisects angle S and RQ bisects R, angle ARN = angle MRS and angle MSR = angle MSC.)
Now,
By A.S.P of triangle RMS,
angle MSR + angle MRS + angle RMS = 180°-----(2)
substituting eq.(1) in eq.(2),
(MSR + MRS )+ RMS= 180°
90° + angle RMS= 180°
angle RMS = 180° - 90°
angle RMS = 90°
similarly, by A.S.P of triangle RNS,
angle RNS + angle NRS + angle RSN = 180°
RNS + (NRS + RSN)= 180°
RNS + 90°=180°
RNS= 180° - 90°
RNS = 90°
thereby,
angle RMS = RNS = 90°
So, RM perpendicular to SN and SN perpendicular to RM.
HENCE PROVED.