Question

Solve the following Activity​

Answer 1

Answer:

$\huge \sf \underbrace{ \: Proof \: }$

$\sf \implies \angle \: EFG = \angle \: FGH... \fbox {( \sf \green{ alternate \: angle)}}...1$

$\sf \implies \angle \: EFG \: = \angle \fbox {\sf \green{ FGH}}...( \sf \: inscribed \: angle \: theorem)...2$

$\sf \implies \angle\fbox{ \green{ EFG }} = \angle {\sf { FGH}}...( \sf \: inscribed \: angle \: theorem)..3$

$\sf \implies \therefore \: m(arcEG) \: ≅ \fbox{ \sf \green {m(arcFH)}}...(from \: 1,2,3)$

$\sf \implies \therefore \: chord \: EG≅chord \: FH$

$\sf ....corresponding \:chord \: of \: the \: congruent \: arcs .$

Answer 2

Proof

$\sf \implies \angle \: EFG = \angle \: FGH... \fbox {( \sf \green{ alternate \: angle)}}...1⟹∠EFG=∠FGH...$

( alternate angle)

1

$\sf \implies \angle \: EFG \: = \angle \fbox {\sf \green{ FGH}}...( \sf \: inscribed \: angle \: theorem)2⟹∠EFG=∠FGH$

(inscribedangletheorem) 2

$\sf \implies \angle\fbox{ \green{ EFG }} = \angle {\sf { FGH}}...( \sf \: inscribed \: angle \: theorem)..3⟹∠EFG$

=∠FGH (inscribedangletheorem) 3

$\sf \implies \therefore \: m(arcEG) \: ≅ \fbox{ \sf \green {m(arcFH)}}...(from \: 1,2,3)⟹∴m(arcEG)≅ </p><p> m(arcFH)$

...(from1,2,3)

$\sf \implies \therefore \: chord \: EG≅chord \: FH⟹∴chordEG≅chordFH</p><p>\sf ....corresponding \:chord \: of \: the \: congruent \:$

$arcs \: corresponding \: chord \: of \: the \: congruentarcs$