C-07-Q062mediumsingle_mcq

From $\angle ABO = \angle CDO$ being alternate angles, it follows that—

a $AB \perp CD$
b $AB$ and $CD$ are parallel and equal
c $AC = BD$
d $\triangle ABC$ is right-angled

ব্যাখ্যা

Equal alternate angles ( $\angle ABO = \angle CDO$ ) force $AB \parallel CD$ , and the congruent triangles give $AB = CD$ . Hence $AB$ and $CD$ are both parallel and equal, which is exactly the condition for the opposite sides of a parallelogram.

এই অধ্যায়ের আরও প্রশ্ন

C-07-Q063In Construction 4, the included angle between the diagonals is—
C-07-Q064In Construction 5, on ray the segment is cut equal to—
C-07-Q065The arcs at and are drawn with radii respectively—
C-07-Q066The point of intersection of these arcs gives—
C-07-Q067To complete the parallelogram, is cut from equal to—
C-07-Q068In Construction 5, is justified by—
C-07-Q069In Example 3 the parallel sides satisfy—
C-07-Q070From ray , the line segment is cut equal to—
C-07-Q071From line segment , is cut equal to—
C-07-Q072The line through parallel to meets ray at—

Practical Geometry — সব প্রশ্ন দেখুন (150টি) →

From $\angle ABO = \angle CDO$ being alternate angles, it follows that—

AB \perp CD

AB

and

CD

are parallel and equal

AC = BD

\triangle ABC

is right-angled

From ∠ABO=∠CDO\angle ABO = \angle CDO∠ABO=∠CDO being alternate angles, it follows that—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

From ∠ABO=∠CDO\angle ABO = \angle CDO∠ABO=∠CDO being alternate angles, it follows that—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

From $\angle ABO = \angle CDO$ being alternate angles, it follows that—

From $\angle ABO = \angle CDO$ being alternate angles, it follows that—