Question

9. 3) Prove that two opposite vertices of a
parallelogram are equidistant from the
diagonal not containing these vertices.​

Answer 1

Hence Proved

Step-by-step explanation:

We will prove this by taking the perpendicular distances of the opposite vertices from the diagonal not containing them.

Construction: Draw a perpendicular from B and D such that they meet AC at N and P respectively.

To prove, BP = DN

In ΔADC and Δ ABC,

AD=BC(Opposite sides of a parallelogram)

AB=DC (Opposite sides of a parallelogram)

AC =AC (Common)

Therefore,ΔADC ≅ ΔCBA (By SSS Concruency )

⇒ Area of ΔADC = Area of Δ CBA (Since,Area of congruent triangles are equal)

We know that,$Area of triangle = \frac{1}{2}\times base\times Height$

⇒$\frac{1}{2}\times AC\times DN = \frac{1}{2}\times AC\times BP$

⇒ DN = BP

Therefore,The perpendicular distance of opposite vertices from the diagonal of a parallelogram ,not containing  them is equal.

Answer 2

Step-by-step explanation:

HOPE SO YOU UNDERSTAND MY ANSWER