C-14-Q031mediumsingle_mcq

The heights of $\triangle ADE$ and $\triangle BDE$ (with the common vertex pattern of the proof) are —

aequal
bunequal
czero
dparallel

ব্যাখ্যা

Triangles $ADE$ and $BDE$ have their bases $AD$ and $DB$ lying on the same straight line $AB$ , with the common opposite vertex $E$ . The perpendicular distance from $E$ to line $AB$ is therefore the same for both, making their heights equal.

এই অধ্যায়ের আরও প্রশ্ন

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 —
C-14-Q035In the proof of Theorem 29, because they are on the same base and —
C-14-Q036If on with , , and on with , , then is —
C-14-Q037The internal bisector of an angle of a triangle divides the opposite side in the ratio of —
C-14-Q038In , bisects and meets at . Then —
C-14-Q039In the proof of Theorem 30, the auxiliary line is drawn —
C-14-Q040In the proof, because they are —
C-14-Q041In the proof, because they are —

The heights of $\triangle ADE$ and $\triangle BDE$ (with the common vertex pattern of the proof) are —

aequal

bunequal

czero

dparallel

The heights of △ADE\triangle ADE△ADE and △BDE\triangle BDE△BDE (with the common vertex pattern of the proof) are —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The heights of △ADE\triangle ADE△ADE and △BDE\triangle BDE△BDE (with the common vertex pattern of the proof) are —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The heights of $\triangle ADE$ and $\triangle BDE$ (with the common vertex pattern of the proof) are —

The heights of $\triangle ADE$ and $\triangle BDE$ (with the common vertex pattern of the proof) are —