C-14-Q092mediumsingle_mcq

With $DE \parallel BC$ in $\triangle ABC$ , the correct relation is —

a $\dfrac{AD}{BD} = \dfrac{AE}{EC}$
b $\dfrac{AD}{BD} = \dfrac{CE}{AB}$
c $\dfrac{AD}{AE} = \dfrac{BD}{CE}$ only
d $\dfrac{AD}{AB} = \dfrac{EC}{AC}$

ব্যাখ্যা

A line parallel to one side of a triangle divides the other two sides proportionally (Basic Proportionality / Thales' theorem). Hence $\frac{AD}{BD} = \frac{AE}{EC}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q093With , the area ratio is —
C-14-Q094In Exercise 14.2 Q1, the true statements are —
C-14-Q095In , with , . The correct relation is —
C-14-Q096If each of two triangles is similar to a third triangle, then they are —
C-14-Q097Two right-angled triangles with one pair of acute angles equal are —
C-14-Q098The perpendicular from the right-angle vertex of a right triangle splits it into two triangles each similar to
C-14-Q099In Exercise 14.2 Q11, when equals —
C-14-Q100In Q12, two similar triangles have cm, cm, and area sq cm. Then area
C-14-Q101In a parallelogram , a line through meets at and at . Then
C-14-Q102If two equiangular triangles and have heights , from matching vertices, then —

With $DE \parallel BC$ in $\triangle ABC$ , the correct relation is —

\dfrac{AD}{BD} = \dfrac{AE}{EC}

\dfrac{AD}{BD} = \dfrac{CE}{AB}

\dfrac{AD}{AE} = \dfrac{BD}{CE}

only

\dfrac{AD}{AB} = \dfrac{EC}{AC}

ব্যাখ্যা

A line parallel to one side of a triangle divides the other two sides proportionally (Basic Proportionality / Thales' theorem). Hence $\frac{AD}{BD} = \frac{AE}{EC}$ .

With DE∥BCDE \parallel BCDE∥BC in △ABC\triangle ABC△ABC, the correct relation is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

With DE∥BCDE \parallel BCDE∥BC in △ABC\triangle ABC△ABC, the correct relation is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

With $DE \parallel BC$ in $\triangle ABC$ , the correct relation is —

With $DE \parallel BC$ in $\triangle ABC$ , the correct relation is —