Question

.......plzzz solve it frnds .....​

Answer 1

Answer:

Step-by-step explanation:

Let R be the centre of the circle.

Let AB be a straight line passing through R and bisecting a chord CD

so that BC=BD

Join RC and RD

In triangle RBC and triangle RBD :

RB=RB (common)

BC=BD(given)

RC=RD(radius of a circle)

By SSS(side side side) congruency, RBC is congruent to RBD

By Cpct, angle RBC = angle RBD

CD is a straight chord

RBC+RBD=180(Linear pair)

2RBC=180 ( since RBC=RBD)

2RBD=180

RBC=RBD=90

AB is a straight line

ABC=RBC=90

so , AB is perpendicular to CD

Hence,

if any straight line passing through the centre of a circle bisects any chords which is not a diameter, then the straight line will be perpendicular on that chord.