C-14-Q035mediumsingle_mcq

In the proof of Theorem 29, $\triangle BDE = \triangle DEC$ because they are on the same base $DE$ and —

abetween the same pair of parallel lines
bshare the third vertex
ccongruent by SAS
dhave equal medians

ব্যাখ্যা

Triangles $BDE$ and $DEC$ stand on the common base $DE$ , and since $DE \parallel BC$ , their opposite vertices $B$ and $C$ lie on a line parallel to $DE$ . Triangles on the same base and between the same parallels have equal area.

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q036If on with , , and on with , , then is —
C-14-Q037The internal bisector of an angle of a triangle divides the opposite side in the ratio of —
C-14-Q038In , bisects and meets at . Then —
C-14-Q039In the proof of Theorem 30, the auxiliary line is drawn —
C-14-Q040In the proof, because they are —
C-14-Q041In the proof, because they are —
C-14-Q042In the proof, because is —
C-14-Q043In , , , . bisects , meets at . Then
C-14-Q044In , , , . If bisects ,
C-14-Q045In , , , . If bisects ,

In the proof of Theorem 29, $\triangle BDE = \triangle DEC$ because they are on the same base $DE$ and —

abetween the same pair of parallel lines

bshare the third vertex

ccongruent by SAS

dhave equal medians

In the proof of Theorem 29, △BDE=△DEC\triangle BDE = \triangle DEC△BDE=△DEC because they are on the same base DEDEDE and —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 29, △BDE=△DEC\triangle BDE = \triangle DEC△BDE=△DEC because they are on the same base DEDEDE and —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 29, $\triangle BDE = \triangle DEC$ because they are on the same base $DE$ and —

In the proof of Theorem 29, $\triangle BDE = \triangle DEC$ because they are on the same base $DE$ and —