C-14-Q048mediumsingle_mcq

In the proof of Theorem 31, $BA:AE = BD:DC$ follows from —

aTheorem 28
bTheorem 32
cSAS congruence
dPythagoras' theorem

ব্যাখ্যা

With $CE \parallel DA$ in triangle $BCE$ , the Basic Proportionality Theorem (Theorem 28) applies to the sides cut by the parallel line, giving $BA:AE = BD:DC$ .

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q049In Theorem 31, is established from —
C-14-Q050Bisectors of two base angles of a triangle meet the opposite sides at and . If base, the triangle is —
C-14-Q051If two transversals cut a set of parallel lines, the matching intercepted segments are —
C-14-Q052The diagonals of a trapezium intersect each other —
C-14-Q053The line joining the midpoints of the oblique sides of a trapezium is —
C-14-Q054In , medians , meet at . A line through parallel to meets at . Then
C-14-Q055In , on , on . Then
C-14-Q056Congruence is a special case of —
C-14-Q057Two congruent figures are —
C-14-Q058Two similar figures are —

Ratio, Similarity and Symmetry — সব প্রশ্ন দেখুন (160টি) →

In the proof of Theorem 31, BA:AE=BD:DCBA:AE = BD:DCBA:AE=BD:DC follows from —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 31, BA:AE=BD:DCBA:AE = BD:DCBA:AE=BD:DC follows from —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 31, $BA:AE = BD:DC$ follows from —

In the proof of Theorem 31, $BA:AE = BD:DC$ follows from —