C-14-Q024mediumsingle_mcq

The proof of Theorem 28 uses the fact that $\triangle BDE$ and $\triangle DEC$ have —

aequal heights
bequal bases and lie between the same parallels
cequal angles only
dequal sides

ব্যাখ্যা

$\triangle BDE$ and $\triangle DEC$ share the same base $DE$ and lie between the parallels $DE$ and $BC$ , so they have equal areas. This equal-area relation is the key step in proving the proportionality theorem.

এই অধ্যায়ের আরও প্রশ্ন

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 —
C-14-Q034In , if where , , then —

The proof of Theorem 28 uses the fact that $\triangle BDE$ and $\triangle DEC$ have —

aequal heights

bequal bases and lie between the same parallels

cequal angles only

dequal sides

The proof of Theorem 28 uses the fact that △BDE\triangle BDE△BDE and △DEC\triangle DEC△DEC have —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The proof of Theorem 28 uses the fact that △BDE\triangle BDE△BDE and △DEC\triangle DEC△DEC have —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The proof of Theorem 28 uses the fact that $\triangle BDE$ and $\triangle DEC$ have —

The proof of Theorem 28 uses the fact that $\triangle BDE$ and $\triangle DEC$ have —