`A B C D` is a quadrilateral. A line through `D ,` parallel to `A C ,` meets `B C` produced in `P` as shown in figure. Prove that `a r (triangle A B P)`=a r (Quad.` A B C D)`.

In Figure, A B C D E is a pentagon. A line through B parallel to A C meets D C produced at F . Show that: (i) a r( triangle A C B)=a r(triangle A C F) (ii) a r( quadrilateral A E D F)=a r(A B C D E)

In Figure, A B C D E\ is a pentagon. A line through B parallel to A C meets D C produced at F . Show that: (i) a r\ ( A C B)=a r\ ( A C F) (ii) a r\ (A E D F)=a r\ (A B C D E)

The side A B of a parallelogram A B C D is produced to any point Pdot A line through A parallel to C P meets C B produced in Q and the parallelogram P B Q R completed. Show that a r(llgm A B C D)=a r(llgm B P R Q)dot CONSTRUCTION : Join A C and PQ. TO PROVE : a r(llgm A B C D)=a r(llgm B P R Q)

The side A B of a parallelogram A B C D is produced to any point Pdot A line through A and parallel to C P meets C B produced at Q and then parallelogram P B Q R is completed as shown in Figure. Show that a r(^(gm)A B C D)=a r ^(gm)P B Q R)dot

The side A B of a parallelogram A B C D is produced to any point Pdot A line through A parallel to C P meets C B produced in Q and the parallelogram P B Q R completed. Show that a r(^(gm)A B C D)=a r(^(gm)B P R Q)dot CONSTRUCTION : Join A C and PQ. TO PROVE : a r(^(gm)A B C D)=a r(^(gm)B P R Q)

A B C D is a trapezium with A B || D C . A line parallel to A C intersects A B at X and B C at Y . Prove that a r( triangle A D X)=a r(triangle A C Y) .

The sides B A and D C of a quadrilateral A B C D are produced as shown in Figure. Prove that a+b=x+ydot Figure

In Figure, A B C D is a quadrilateral and B E || A C and also B E meets D C produced at E . Show that area of triangle A D E is equal to the area of the quadrilateral A B C D .

A B C D is a cyclic quadrilateral in which B A\ a n d\ C D when produced meet in E\ a n d\ E A=E Ddot Prove that: (i) A D is p a r a l l e l t o B C (ii) E B=E C

A B C D is a parallelogram whose diagonals intersect at O . If P is any point on B O , prove that: a r\ (triangle \ A B P)=\ a r\ (triangle \ C B P)