C-07-Q018mediumsingle_mcq

In $\triangle ACD$ , since $\angle ADC = \angle ACD$ , it follows that—

a $AC = BC$
b $AC = AD$
c $AD = CD$
d $AB = AC$

ব্যাখ্যা

Equal base angles in $\triangle ACD$ ( $\angle ADC = \angle ACD$ ) make it an isosceles triangle, so the sides opposite those equal angles are equal, giving $AC = AD$ .

এই অধ্যায়ের আরও প্রশ্ন

C-07-Q019In the constructed triangle , equals—
C-07-Q020In the alternate method of Construction 1, line segment is bisected by—
C-07-Q021In and , the included angle equals—
C-07-Q022The congruence is established by—
C-07-Q023After joining and in the alternate method, equals—
C-07-Q024In Construction 2 the difference is cut on ray as line segment—
C-07-Q025The angle constructed at in Construction 2 equals—
C-07-Q026In the resulting triangle , the difference equals—
C-07-Q027If the given base-adjacent angle is NOT acute, Construction 2 is—
C-07-Q028In Construction 2, because in —

Practical Geometry — সব প্রশ্ন দেখুন (150টি) →

In $\triangle ACD$ , since $\angle ADC = \angle ACD$ , it follows that—

AC = BC

AC = AD

AD = CD

AB = AC

ব্যাখ্যা

Equal base angles in $\triangle ACD$ ( $\angle ADC = \angle ACD$ ) make it an isosceles triangle, so the sides opposite those equal angles are equal, giving $AC = AD$ .

In △ACD\triangle ACD△ACD, since ∠ADC=∠ACD\angle ADC = \angle ACD∠ADC=∠ACD, it follows that—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △ACD\triangle ACD△ACD, since ∠ADC=∠ACD\angle ADC = \angle ACD∠ADC=∠ACD, it follows that—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle ACD$ , since $\angle ADC = \angle ACD$ , it follows that—

In $\triangle ACD$ , since $\angle ADC = \angle ACD$ , it follows that—