Question

In the figure given beliw, ABCD is a kite in which BC=CD , AB= AD. (1) angle EFG = 90 degree. (2) the line drawn through G and parallel to FE bisects DA. ​

Answer 1

Answer:

join EG

bisect EG

stretch the bisector of EG

Answer 2

Step-by-step explanation:

It is given that,

It is given that,ABCD is a kite in which BC = CD, AB = AD;

It is given that,ABCD is a kite in which BC = CD, AB = AD;E, F, G are midpoints of CD, BC and AB

${\huge{\red{\text{To prove : ∠EFG = 90°}}}}$

Construction : Join AC and BD ,Construct GH through G paralled to FE.

Proof :We know that,

Diagonals of a kite intersect at right angles.

∴${\boxed{\orange{\text{ ∠MON = 90° }}}}$ ...(1)

In △BCD,

E and F are midpoints of CD and BC

 [ By Basic Proportionality Theorem ]

⇒ EF || DB and EF = 1/2DB...(2)

Now, EF || DB ⇒ MF || ON

Similarly, FG || CA ⇒ FN || MO

Therefore, In □MFNO,

MF || ON, FN || MO and ∠MON = 90°

⇒${\boxed{\underline{\blue{\text{ □MFNO is a square. }}}}}$

∴${\huge{\boxed{\pink{\text{∠EFG=90°}}}}}$