LF+FEED2 (ABBC+AC)T17. ABCD is a parallelogram in whelogram in which b

1.	LF+FEED2 (ABBC+AC)T17. ABCD is a parallelogram in whelogram in which bisectors of ZA and ZC meet the diagonalBD at P and Q respectirvely Prove that PCOA is a parallelogram.IS. ABCD is a parallelogram and EF BD. R is mid point of EF. Prove thatE C
Answer» Since diagonal of a square bisects the vertex and BD is the diagonal of square ABCD CBD=CBD=45° Given :EF \|\| BD ⇒ ∠CEF = ∠CBD = 45° and ∠CEF = ∠CDB = 45° (corresponding angles) ⇒ ∠CEF = ∠CFE ⇒ CE = CF (sides opposite of equal angles are equal) .......(1) Now BC = CD (Sides of square) ........(2) Subtracting (1) from (2) we get ⇒ BC – CE = CD – CF ⇒ BE = DF

LF+FEED2 (ABBC+AC)T17. ABCD is a parallelogram in whelogram in which bisectors of ZA and ZC meet the diagonalBD at P and Q respectirvely Prove that PCOA is a parallelogram.IS. ABCD is a parallelogram and EF BD. R is mid point of EF. Prove thatE C

Answer»

Since diagonal of a square bisects the vertex and BD is the diagonal of square ABCD

CBD=CBD=45°

Given :EF || BD

⇒ ∠CEF = ∠CBD = 45° and ∠CEF = ∠CDB = 45° (corresponding angles)

⇒ ∠CEF = ∠CFE

⇒ CE = CF (sides opposite of equal angles are equal) .......(1)

Now BC = CD (Sides of square) ........(2)

Subtracting (1) from (2) we get

⇒ BC – CE = CD – CF

⇒ BE = DF