To get a definite picture, we consider the situation drawn in the figure 1. In t

Question

To get a definite picture, we consider the situation drawn in the figure 1. In that case, the two given lines intersect in point O, and the points mentioned above occur on the two lines in the order O*B*C*F, and O*A*D*E. We need to compare the two angles and . The angle is exterior angle in the circular quadrilateral ABCD. By the remark above, it is congruent to the opposite interior angle: Now the second angle is exterior angle in the second circular quadrilateral CDEF.Hence it is congruences to the opposite interior angle: From these two congruences, we conclude . Hence, by Euclid I.27, the lines AB and FE are parallel. Question. Explain how the proof has to be modified in the case drawn in the figure 2. Answer: ???

Explanation / Answer

Question Details:YOUREXCELLENTLY DRAWN DIAGRAM IS
UNFORTUNATELY NOT SHOWING UP IN DRAWING EDITOR
SO PLEASE IMAGINE THE FOLLOWING CONSTRUCTION
JOIN BD AND CA.
LET
ANGLE BDC=X
AND
ANGLE BCA=Y
BCAD IS A CYCLIC QUADRILATERAL
ANGLE ODB IS EXTERIOR ANGLE = ANGLE BCA = Y
ANGLE ODB = ANGLE ODC - ANGLE BDC=95-X
HENCE
95-X=Y............X+Y=95.....................1
BC IS A COMMON CHORD AND ANGLES BDC AND BAC
ARE ANGLES IN THE SAME SEGMENT . HENCE
ANGLE BDC = ANGLE BAC=X
IN TRIANGLE ABC
ANGLE ABC = 180-(ANGLE BAC+ANGLE BCA) =180-(X+Y)
OBC BEING A ST.LINE
ANGLE OBA = 180-ANGLE ABC =180-[180-(X+Y)]=X+Y................2
BUT FROM EQN. 1 ABOVE
X+Y=95
HENCE ANGLE OBA = 95 =ANGLE EFB
EF AND AB ARE 2 LINES OBF IS TRANSVERSAL.
ANGLE OBA=95=ANGLE EFB ....
CORRESPONDING ANGLES ARE EQUAL
HENCE AB || EF....PROVED

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.