Question

prove that diagonal of //gm are equal

Answer 1

Let ACDB be a parallelogram, and let BC be a diameter.

By definition of parallelogram, AB∥CD , and BC intersects both.

So by Parallelism implies Equal Alternate Interior Angles:

∠ABC=∠BCD

AC∥BD , and BC intersects both.

∠ACB=∠CBD

So △ABC and △DCB have two angles equal, and the side BC in common.

So by Triangle Angle-Side-Angle Equality:

△ABC=△DCB

So AC=BD and AB=CD .

Also, we have that ∠BAC=∠BDC .

So we have ∠ACB=∠CBD and ∠ABC=∠BCD .

So by Common Notion 2:

∠ACB+∠BCD=∠ABC+∠CBD

So ∠ACD=∠ABD .

So we have shown that opposite sides and angles are equal to each other.

Now note that AB=CD , and BC is common, and ∠ABC=∠BCD .

So by Triangle Side-Angle-Side Equality:

△ABC=△BCD

So BC bisects the parallelogram.

Similarly, AD also bisects the parallelogram.

Answer 2

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º.
In ΔABC and ΔDCB
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
∴ ΔABC ≅ ΔDCB (By SSS Congruence rule)
⇒ ∠ABC = ∠DCB
It is known that the sum of the measures of angles on the same side of transversal is 180º.
∠ABC + ∠DCB = 180º (AB || CD)
⇒ ∠ABC + ∠ABC = 180º
⇒ 2∠ABC = 180º
⇒ ∠ABC = 90º
hence proved opposite sides r parallel