C-14-Q023mediumsingle_mcq

In $\triangle ABC$ , $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ . Then —

a $AD:DB = AE:EC$
b $AD:AB = AE:AC$ only
c $AD = AE$
d $DB = EC$

ব্যাখ্যা

Since $DE\parallel BC$ , the Basic Proportionality Theorem divides $AB$ and $AC$ in equal ratios, giving $AD:DB=AE:EC$ . The segments need not be equal in length—only proportional.

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q024The proof of Theorem 28 uses the fact that and have —
C-14-Q025Corollary 1 of Theorem 28 gives —
C-14-Q026Corollary 2 of Theorem 28 says: a line through the midpoint of one side parallel to another side —
C-14-Q027In , . If , , , then
C-14-Q028In , . If , , , then
C-14-Q029In , , . Then
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 —

In $\triangle ABC$ , $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ . Then —

AD:DB = AE:EC

AD:AB = AE:AC

only

AD = AE

DB = EC

ব্যাখ্যা

Since $DE\parallel BC$ , the Basic Proportionality Theorem divides $AB$ and $AC$ in equal ratios, giving $AD:DB=AE:EC$ . The segments need not be equal in length—only proportional.

In △ABC\triangle ABC△ABC, DE∥BCDE \parallel BCDE∥BC with DDD on ABABAB and EEE on ACACAC. Then —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △ABC\triangle ABC△ABC, DE∥BCDE \parallel BCDE∥BC with DDD on ABABAB and EEE on ACACAC. Then —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle ABC$ , $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ . Then —

In $\triangle ABC$ , $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ . Then —