Question

(a) In the Figure (1) given below, $ABCD$ and $ABEF$ are parallelograms. Prove that
(i) $CDFE$ is a parallelogram
(ii) $FD=EC$
(iii) $\Delta AFD=\Delta BEC$.
(b) In the figure (2) given below, $ABCD$ is a parallelogram, $ADEF$ and $AGHB$ are two squares.
Prove that $FG=AC$.

Answer 1

Given $ABCD$ and $ABEF$ are $||$ gms
To prove: (i) $CDEF$ is a parallelogram
(ii) $FD=EC$
(iii) $\Delta AFD=\Delta BEC$
Proof:
(1) $DC||AB$ and $DC=AB\left[ABCDisa||gm\right]$
(2) $FE||AB$ and $FE=AB\left[ABEFisa||gm\right]$
(3) $DC||FE$ and $DC=FE\left[From\right(1\left)and\right(2\left)\right]$
Thus, $CDEF$ is a $||gm$
(4) $CDEF$ is a $||gm$
So, $FD=EC$
(5) In $\Delta AFD$ and $\Delta BEC$, we have $AD=BC$ [Opposite sides of $||gmABCD$ are equal]
$AF=BE$ [Opposite sides of $||gmABEF$ are equal]
$FD=BE$ [From (4)]
Hence, $\Delta AFD\cong \Delta BEC$ by S.S.S axiom of congruency
(b) Given: $ABCD$ is a $||gm,ADEF$ and $AGHB$ are two squares
To prove: $FG=AC$
Proof:
(1) $\angle FAG+{90}^o+{90}^o+\angle BAD={360}^o$
[At a point total angle is ${360}^o$]
$\angle FAG={360}^o-{90}^o-{90}^o-\angle BAD$
$\angle FAG={180}^o-\angle BAD$
(2) $\angle B+\angle BAD={180}^o$ [Adjacent angle in $||gm$ is equal to ${180}^o$]
$\angle B={180}^o-\angle BAD$
(3) $\angle FAG=\angle B$ [From (1) and (2)]
(4) In $\Delta AFG$ and $\Delta ABC$, we have $AF=BC$ [$FADE$ and $ABCD$ both are squares on the same base]
Similarly, $AG=AB$
$\angle FAG=\angle B$ [From (3)]
So, $\Delta AFG\cong \Delta ABC$ by $S.A.S$ axiom of congruency
Hence, By $C.P.C.T$
$FG=AC$