Step 1

Given:

Step 2

We have,

$In\mathrm{\u25b3}FGH{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\mathrm{\u25b3}FJH$,

FG=FJ(Given)

GH=JH(Given)

FH=FH(Common side)

So, by SSS congruence criteria

$\mathrm{\u25b3}FGH\stackrel{\sim}{=}\mathrm{\u25b3}FJH$...(i)

A.

$In\mathrm{\u25b3}FGK{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\mathrm{\u25b3}FJK$,

FG=FJ(Given)

$\mathrm{\angle}GFK=\mathrm{\angle}JFK$ [Using(i)]

FK=FK(Common side)

So, by SAS congruence criteria

$\mathrm{\u25b3}FGK\stackrel{\sim}{=}\mathrm{\u25b3}FJK$
B.Similarity, using(i) and A, we get

$In\mathrm{\u25b3}GKH{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\mathrm{\u25b3}JKH$,

GK=JK[Using(A)]

GH=JH(Given)

KH=KH(Common side)

So, by SSS congruence criteria

$\mathrm{\u25b3}GKH\stackrel{\sim}{=}\mathrm{\u25b3}JKH$
Hence,

$\mathrm{\angle}GKH\stackrel{\sim}{=}\mathrm{\angle}JKH$
D.Using(i),

$\mathrm{\angle}GFH\stackrel{\sim}{=}\mathrm{\angle}JFH$
E.Already proved in B,

$\mathrm{\u25b3}GKH\stackrel{\sim}{=}\mathrm{\u25b3}JKH$
C & F. These two points are False.

Step 3

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