C-14-Q029mediumsingle_mcq

In $\triangle ABC$ , $DE \parallel BC$ , $AD:DB = 2:3$ . Then $AE:EC = ?$

a $3:2$
b $2:3$
c $2:5$
d $5:2$

ব্যাখ্যা

Because $DE\parallel BC$ , the line divides both sides in the same ratio, so $AE:EC=AD:DB=2:3$ . The corresponding segments are directly proportional.

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q030In , , , , . Then
C-14-Q031The heights of and (with the common vertex pattern of the proof) are —
C-14-Q032If a line segment divides two sides of a triangle (or their produced sections) proportionally, it is —
C-14-Q033Theorem 29 is the converse of —
C-14-Q034In , if where , , then —
C-14-Q035In the proof of Theorem 29, because they are on the same base and —
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 —

Ratio, Similarity and Symmetry — সব প্রশ্ন দেখুন (160টি) →

In △ABC\triangle ABC△ABC, DE∥BCDE \parallel BCDE∥BC, AD:DB=2:3AD:DB = 2:3AD:DB=2:3. Then AE:EC=?AE:EC = ?AE:EC=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △ABC\triangle ABC△ABC, DE∥BCDE \parallel BCDE∥BC, AD:DB=2:3AD:DB = 2:3AD:DB=2:3. Then AE:EC=?AE:EC = ?AE:EC=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle ABC$ , $DE \parallel BC$ , $AD:DB = 2:3$ . Then $AE:EC = ?$

In $\triangle ABC$ , $DE \parallel BC$ , $AD:DB = 2:3$ . Then $AE:EC = ?$