Question-9
Q.9. In the Fig., given below ABCD is a parallelogram and if , prove that .
In △MPQ,
AB / PQ = MA / MP
=> CD / PQ = MA / MP [ ... AB = CD (sides of a | |gm) ]
But, In △RPQ,
CD / PQ = RD / RP
Therefore, MA / MP = RD / RP
=> AD | | MR
=> BC | | MR [... AD | | BC (sides of a | |gm) ]
This was just a hint on how to solve this question. You need to do the necessary steps ( like proving triangles similar ) in between. I am prretty sure you will be able to do that.
Hope. . it helps you
AB / PQ = MA / MP
=> CD / PQ = MA / MP [ ... AB = CD (sides of a | |gm) ]
But, In △RPQ,
CD / PQ = RD / RP
Therefore, MA / MP = RD / RP
=> AD | | MR
=> BC | | MR [... AD | | BC (sides of a | |gm) ]
This was just a hint on how to solve this question. You need to do the necessary steps ( like proving triangles similar ) in between. I am prretty sure you will be able to do that.
Hope. . it helps you