# M is a point on t

(i). Given: ABCD is a parallelogram.

To Prove:

Proof: In ∆DMC and ∆NMB,

DMC = NMB [ they are vertically opposite angles]

DCM = NBM [ they are alternate angles]

CDM = MNB [ they are alternate angles]

By AAA-similarity, we can say

∆DMC ∆NMB

So, from similarity of the triangle, we can say

Hence, proved.

(ii). Given: ABCD is a parallelogram.

To Prove:

Proof: As we have already derived

Add 1 on both sides of the equation, we get

[ ABCD is a parallelogram and a parallelogram’s opposite sides are always equal DC = AB]

Hence, proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

In Fig. 4.243, thRD Sharma - Mathematics

In Fig. 4.246, ifRD Sharma - Mathematics

In Fig. 4.244, ifRD Sharma - Mathematics

If <img widRD Sharma - Mathematics

If <img widRD Sharma - Mathematics

If in <img RD Sharma - Mathematics

In a <img wRD Sharma - Mathematics

If <img widRD Sharma - Mathematics

In <img widRD Sharma - Mathematics