C-14-Q025mediumsingle_mcq

Corollary 1 of Theorem 28 gives —

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

ব্যাখ্যা

Starting from $AD:DB=AE:EC$ and applying componendo (adding the denominator's whole part), the corollary gives $\frac{AB}{AD}=\frac{AC}{AE}$ , since $AB=AD+DB$ and $AC=AE+EC$ . This relates the full sides to their upper segments.

এই অধ্যায়ের আরও প্রশ্ন

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 —
C-14-Q034In , if where , , then —
C-14-Q035In the proof of Theorem 29, because they are on the same base and —

Corollary 1 of Theorem 28 gives —

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

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

\dfrac{AB}{DB} = \dfrac{AE}{AC}

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