Question

ABCD Is a cyclic quadrilateral with diagonals AC and BD intersecting at m. A line through m bisects ab. Prove that the line is perpendicular to the other side(cd)

Answer 1

solutions :

Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at M at right-angle. MP, bisecting side CD when produced, meets AB is L

To prove:

/_ PLB= 90⁰ DP=CP :. PM=PC=PD

[ :• the median bisecting the hypotenuse of a right-angle is half of the hypotenuse]

Now in CPM,MP =CP=> /_1 = /_2 but /_1 = /_3

[ Angle in the same segment of a circle] and /_2 /_4 (vertically opposite angle) :• /_3=/_4

But /_4+/_5=90°(given) => /_3+/_5 =90°

Now in MBL, /_ MBL+/_BML+/_MLB=180°Or 90°+/_MLB=180°

Hence,/_MLB=90°=>/_LPB=90°

Answer 2

Step-by-step explanation:

Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at M at right-angle. MP, bisecting side CD when produced, meets AB is L

To prove:

/_ PLB= 90⁰ DP=CP :. PM=PC=PD

[ :• the median bisecting the hypotenuse of a right-angle is half of the hypotenuse]

Now in CPM,MP =CP=> /_1 = /_2 but /_1 = /_3

[ Angle in the same segment of a circle] and /_2 /_4 (vertically opposite angle) :• /_3=/_4

But /_4+/_5=90°(given) => /_3+/_5 =90°

Now in MBL, /_ MBL+/_BML+/_MLB=180°Or 90°+/_MLB=180°

Hence,/_MLB=90°=>/_LPB=90°