Question

In Fig-7.26,can you use ASA Congruence rule and conclude that ΔAOC≅ΔBOD?​

Answer 1

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In tri. AOC and tri. BOD,

<ACO = <ODB = 70°

<CAO = < DBO =80°

AC = BD = 3 cm

So, tri.AOC is congruent to tri.BOD by ASA congruency rule.

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Answer 2

$\sf \pmb{☞︎ln \: the \: two \: triangles \: AOC \: and \: BOD} \\ \sf \gray{ \angle \: C = \angle \: D(each \: 70 \degree)}$

$\sf \gray{ \angle \: AOC = \angle \: BOD = 30 \degree(vertically \: opposite \: angles)}$

$\sf \gray{ \angle \:A \: of \: \triangle AOC \: = 180 \degree - (70 \degree + 30 \degree) = 80 \degree}$

$\large \frak { \: \: \: \: \: \: \: (using \: angle \: sum \: property \: of \: a \: triangle.)}$

$\sf \gray{ \angle \:B \: of \: \triangle \:BOD = 180 \degree - (70 \degree + 30 \degree) = 80 \degree)}$

$\sf \gray{ \angle \: A \: and \: \angle \: B \: } \\ \sf \gray{AC \: = BD} \\ \sf \gray{ \angle \:C = \angle \: D }$

$\sf \gray{side \: AC \: is \: between \: \angle \: A \: and \: \angle \: C \: } \\ \sf \gray{side \: BD \: is \: between \: \angle \: B \: and \: \angle \:D }$

$\small\sf \pmb{so \: by \: } \huge \sf{asa} \small\sf \pmb{congruence \: rule \: \triangle \: AOC≅ \triangle \:BOD }$

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