C-14-Q101mediumsingle_mcq

In a parallelogram $ABCD$ , a line through $A$ meets $BC$ at $M$ and $DC$ at $N$ . Then $BM \times DN = ?$

avariable
ba constant
czero
d $AB \cdot CD$

ব্যাখ্যা

Triangles $ABM$ and $NDA$ formed by the transversal are similar (equal angles from the parallel sides), giving $\frac{BM}{AB} = \frac{AD}{DN}$ . Cross-multiplying yields $BM \times DN = AB \times AD$ , which is fixed by the parallelogram, hence a constant.

এই অধ্যায়ের আরও প্রশ্ন

avariable

ba constant

czero

AB \cdot CD

In a parallelogram ABCDABCDABCD, a line through AAA meets BCBCBC at MMM and DCDCDC at NNN. Then BM×DN=?BM \times DN = ?BM×DN=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In a parallelogram ABCDABCDABCD, a line through AAA meets BCBCBC at MMM and DCDCDC at NNN. Then BM×DN=?BM \times DN = ?BM×DN=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In a parallelogram $ABCD$ , a line through $A$ meets $BC$ at $M$ and $DC$ at $N$ . Then $BM \times DN = ?$

In a parallelogram $ABCD$ , a line through $A$ meets $BC$ at $M$ and $DC$ at $N$ . Then $BM \times DN = ?$