C-14-Q034mediumsingle_mcq

In $\triangle ABC$ , if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ where $D \in AB$ , $E \in AC$ , then —

a $DE \parallel BC$
b $DE \perp BC$
c $DE = BC$
d $DE = \dfrac{1}{2} BC$

ব্যাখ্যা

The condition $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ means $D$ and $E$ divide the two sides in the same ratio. By the converse of the Basic Proportionality Theorem, the line $DE$ must then be parallel to the third side $BC$ .

এই অধ্যায়ের আরও প্রশ্ন

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 —
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 ,

In $\triangle ABC$ , if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ where $D \in AB$ , $E \in AC$ , then —

DE \parallel BC

DE \perp BC

DE = BC

DE = \dfrac{1}{2} BC

In △ABC\triangle ABC△ABC, if ADDB=AEEC\dfrac{AD}{DB} = \dfrac{AE}{EC}DBAD​=ECAE​ where D∈ABD \in ABD∈AB, E∈ACE \in ACE∈AC, then —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △ABC\triangle ABC△ABC, if ADDB=AEEC\dfrac{AD}{DB} = \dfrac{AE}{EC}DBAD​=ECAE​ where D∈ABD \in ABD∈AB, E∈ACE \in ACE∈AC, then —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle ABC$ , if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ where $D \in AB$ , $E \in AC$ , then —

In $\triangle ABC$ , if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ where $D \in AB$ , $E \in AC$ , then —