C-14-Q049mediumsingle_mcq

In Theorem 31, $AE = AC$ is established from —

a $BD:DC = BA:AE$ and $BD:DC = BA:AC$
bPythagoras' theorem
cequal areas
dsimilar triangles

ব্যাখ্যা

The given hypothesis states $BD:DC = BA:AC$ , while the construction yields $BD:DC = BA:AE$ . Equating the two ratios forces $AE = AC$ , which then proves $DA$ bisects the angle.

এই অধ্যায়ের আরও প্রশ্ন

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 —
C-14-Q059Polygons with equal numbers of sides whose angles are sequentially equal are —

In Theorem 31, $AE = AC$ is established from —

BD:DC = BA:AE

and

BD:DC = BA:AC

bPythagoras' theorem

cequal areas

dsimilar triangles

ব্যাখ্যা

The given hypothesis states $BD:DC = BA:AC$ , while the construction yields $BD:DC = BA:AE$ . Equating the two ratios forces $AE = AC$ , which then proves $DA$ bisects the angle.

In Theorem 31, AE=ACAE = ACAE=AC is established from —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In Theorem 31, AE=ACAE = ACAE=AC is established from —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In Theorem 31, $AE = AC$ is established from —

In Theorem 31, $AE = AC$ is established from —