C-14-Q076mediumsingle_mcq

After establishing $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ , we conclude —

a $PQ \parallel BC$
b $PQ \perp BC$
c $PQ = BC$
d $AP = AQ$

ব্যাখ্যা

By the converse of the proportional-segments theorem, when a line cuts two sides of a triangle in the same ratio it must be parallel to the third side, so $PQ \parallel BC$ .

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q077Theorem 34 is the analogue, for similarity, of —
C-14-Q078The ratio of the areas of two similar triangles equals the ratio of the squares of —
C-14-Q079If , then
C-14-Q080with , . The ratio of areas is —
C-14-Q081Two similar triangles have areas in ratio . The ratio of matching sides is —
C-14-Q082with , , area of . Area of
C-14-Q083In the proof of Theorem 35,
C-14-Q084Two similar triangles with matching side ratio have area ratio —
C-14-Q085If areas of two similar triangles are and , the ratio of matching sides is —
C-14-Q086If lies between and with , then divides —

After establishing $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ , we conclude —

PQ \parallel BC

PQ \perp BC

PQ = BC

AP = AQ

ব্যাখ্যা

By the converse of the proportional-segments theorem, when a line cuts two sides of a triangle in the same ratio it must be parallel to the third side, so $PQ \parallel BC$ .

After establishing ABAP=ACAQ\dfrac{AB}{AP} = \dfrac{AC}{AQ}APAB​=AQAC​, we conclude —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

After establishing ABAP=ACAQ\dfrac{AB}{AP} = \dfrac{AC}{AQ}APAB​=AQAC​, we conclude —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

After establishing $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ , we conclude —

After establishing $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ , we conclude —