Select all the true statements. A.△FGK=∼△FJK B.∠GKH=∼JKH C.FG―=∼KG― D.∠GFH=∼∠JFH E.△GKH=∼△JKH F.FG―=∼JK―

Check the solution to "Select all the true statements. A.△FGK=∼△FJK B.∠GKH=∼JKH C.FG―=∼KG―..."

rocedwrp

2020-10-23

Select all the true statements.

△ F G K \tilde{=} △ F J K

∠ G K H \tilde{=} J K H

\overset{―}{F G} \tilde{=} \overset{―}{K G}

∠ G F H \tilde{=} ∠ J F H

△ G K H \tilde{=} △ J K H

\overset{―}{F G} \tilde{=} \overset{―}{J K}

hajavaF

Skilled2020-10-24Added 90 answers

Step 1
Given:

Step 2
We have,

I n △ F G H and △ F J H

,
FG=FJ(Given)
GH=JH(Given)
FH=FH(Common side)
So, by SSS congruence criteria

△ F G H \tilde{=} △ F J H

...(i)
A.

I n △ F G K and △ F J K

,
FG=FJ(Given)

∠ G F K = ∠ J F K

[Using(i)]
FK=FK(Common side)
So, by SAS congruence criteria

△ F G K \tilde{=} △ F J K

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

I n △ G K H and △ J K H

,
GK=JK[Using(A)]
GH=JH(Given)
KH=KH(Common side)
So, by SSS congruence criteria

△ G K H \tilde{=} △ J K H

Hence,

∠ G K H \tilde{=} ∠ J K H

D.Using(i),

∠ G F H \tilde{=} ∠ J F H

E.Already proved in B,

△ G K H \tilde{=} △ J K H

C & F. These two points are False.
Step 3
Hence the true statements are A, B, D, and E.

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