Given:

Step 2

We have,

\(\displaystyle{I}{n}\triangle{F}{G}{H}{\quad\text{and}\quad}\triangle{F}{J}{H}\),

FG=FJ(Given)

GH=JH(Given)

FH=FH(Common side)

So, by SSS congruence criteria

\(\displaystyle\triangle{F}{G}{H}\stackrel{\sim}{=}\triangle{F}{J}{H}\)...(i)

A.\(\displaystyle{I}{n}\triangle{F}{G}{K}{\quad\text{and}\quad}\triangle{F}{J}{K}\),

FG=FJ(Given)

\(\displaystyle\angle{G}{F}{K}=\angle{J}{F}{K}\) [Using(i)]

FK=FK(Common side)

So, by SAS congruence criteria

\(\displaystyle\triangle{F}{G}{K}\stackrel{\sim}{=}\triangle{F}{J}{K}\)

B.Similarity, using(i) and A, we get

\(\displaystyle{I}{n}\triangle{G}{K}{H}{\quad\text{and}\quad}\triangle{J}{K}{H}\),

GK=JK[Using(A)]

GH=JH(Given)

KH=KH(Common side)

So, by SSS congruence criteria

\(\displaystyle\triangle{G}{K}{H}\stackrel{\sim}{=}\triangle{J}{K}{H}\)

Hence, \(\displaystyle\angle{G}{K}{H}\stackrel{\sim}{=}\angle{J}{K}{H}\)

D.Using(i),\(\displaystyle\angle{G}{F}{H}\stackrel{\sim}{=}\angle{J}{F}{H}\)

E.Already proved in B, \(\displaystyle\triangle{G}{K}{H}\stackrel{\sim}{=}\triangle{J}{K}{H}\)

C & F. These two points are False.

Step 3

Hence the true statements are A, B, D, and E.