C-14-Q067mediumsingle_mcq

In the proof of Theorem 32, $\triangle APQ \cong \triangle DEF$ by —

aSSS
bSAS
cAAA
dRHS

ব্যাখ্যা

In the construction, $AP$ and $AQ$ are cut equal to $DE$ and $DF$ , and the included angle $\angle A = \angle D$ . With two sides and the included angle equal, $\triangle APQ \cong \triangle DEF$ by the SAS criterion.

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q068After , we get , hence —
C-14-Q069The relation in the proof follows from —
C-14-Q070If three pairs of corresponding sides of two triangles are proportional, the corresponding angles are —
C-14-Q071In the proof of Theorem 33, by —
C-14-Q072In the proof, and are equiangular because —
C-14-Q073Theorem 33 is the converse of —
C-14-Q074If one angle of a triangle equals an angle of another and the sides adjacent to those angles are proportional,
C-14-Q075In the proof of Theorem 34, by —
C-14-Q076After establishing , we conclude —
C-14-Q077Theorem 34 is the analogue, for similarity, of —

In the proof of Theorem 32, $\triangle APQ \cong \triangle DEF$ by —

aSSS

bSAS

cAAA

dRHS

In the proof of Theorem 32, △APQ≅△DEF\triangle APQ \cong \triangle DEF△APQ≅△DEF by —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 32, △APQ≅△DEF\triangle APQ \cong \triangle DEF△APQ≅△DEF by —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 32, $\triangle APQ \cong \triangle DEF$ by —

In the proof of Theorem 32, $\triangle APQ \cong \triangle DEF$ by —