C-14-Q040mediumsingle_mcq

In the proof, $\angle AEC = \angle BAD$ because they are —

avertically opposite
bcorresponding angles
calternate angles
dco-interior

ব্যাখ্যা

Since $CE \parallel AD$ with $BAE$ acting as the transversal, $\angle AEC$ and $\angle BAD$ are corresponding angles. Corresponding angles formed by a transversal cutting parallel lines are equal.

এই অধ্যায়ের আরও প্রশ্ন

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 ,
C-14-Q046If a side of a triangle is divided internally in the ratio of the other two sides, the segment to the opposite
C-14-Q047In Theorem 31, the construction draws parallel to —
C-14-Q048In the proof of Theorem 31, follows from —
C-14-Q049In Theorem 31, is established from —
C-14-Q050Bisectors of two base angles of a triangle meet the opposite sides at and . If base, the triangle is —

In the proof, $\angle AEC = \angle BAD$ because they are —

avertically opposite

bcorresponding angles

calternate angles

dco-interior

ব্যাখ্যা

Since $CE \parallel AD$ with $BAE$ acting as the transversal, $\angle AEC$ and $\angle BAD$ are corresponding angles. Corresponding angles formed by a transversal cutting parallel lines are equal.

In the proof, ∠AEC=∠BAD\angle AEC = \angle BAD∠AEC=∠BAD because they are —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof, ∠AEC=∠BAD\angle AEC = \angle BAD∠AEC=∠BAD because they are —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof, $\angle AEC = \angle BAD$ because they are —

In the proof, $\angle AEC = \angle BAD$ because they are —