Question

in the adjacent figure, AB is chord of circle with Centre O. CD is diameter perpendicular to AB. show that AD=BD

Answer 1

Given O is the centre and CD is Diameter which is perpendicular to AB. Name the intersecting point of AB and CD as E

hence, CO is also perpendicular to AB

We know that perpendicular from the centre bisects the chord.

Hence, AE = BE

Again, since it is perpendicular,

=> angle AED = angle BED

now in ∆AED and ∆BED

AE = BE (shown above)
angle AED = angle BED (90° each)
ED = ED (common)

Hence, ∆AED is congruent to ∆BED

=> AD = BD (by cpct)

Answer 2

Answer:

Length of AD is equal to the length of BD.

Step-by-step explanation:

Let the point where DC intersects AB is E.

First, we are going to prove that AE = EB.

As OE is perpendicular to AB, by Pythagoras Theorem in triangle OEA,

= >  OA^2 = OE^2 + AE^2

= >  OA^2 - AE^2 = OE^2    ...( i )

Now, by Pythagoras theorem in triangle OEB,

= >  OB^2 = OE^2 + EB^2

= >  OB^2 - EB^2 = OE^2   ...( ii )

Now, comparing ( i ) and ( ii ),

= >  OA^2 - AE^2 = OB^2 - EB^2

= >  OA^2 - OB^2 - AE^2 = - EB^2

As OA and OB are the radius of the given circle, OA = OB & OA^2 = OB^2

So,

= >  OA^2 - OA^2 - AE^2 = - EB^2

= >  - AE^2 = - EB^2

= >  AE^2 = EB^2

= >  AE = EB       ...( iii )

Now, in triangle AED, by Pythagoras Theorem,

= >  AD^2 = AE^2 + ED^2

= >  AD^2 - AE^2 = ED^2   ...( iv )

In triangle DEB, by Pythagoras Theorem,

= >  DB^2 = EB^2 + ED^2

= >  DB^2 - EB^2 = ED^2   ...( v )

Now, comparing ( iv ) and ( v ),

= >  AD^2 - AE^2 = DB^2 - EB^2

= >  AD^2 - AE^2 + EB^2 = DB^2

From ( iii ), AE = EB,

= >  AD^2 - AE^2 + AE^2 = DB^2

= >  AD^2 = BD^2

= >  AD = BD

Hence, proved.