C-14-Q069mediumsingle_mcq

The relation $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ in the proof follows from —

aTheorem 28 (corollary 1)
bPythagoras' theorem
cSAS congruence
dparallelogram law

ব্যাখ্যা

With $PQ \parallel BC$ , the line divides the two sides of $\triangle ABC$ proportionally, which is exactly the result of Theorem 28 (corollary 1). This yields $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q070If three pairs of corresponding sides of two triangles are proportional, the corresponding angles are —
C-14-Q071In the proof of Theorem 33, by —
C-14-Q072In the proof, and are equiangular because —
C-14-Q073Theorem 33 is the converse of —
C-14-Q074If one angle of a triangle equals an angle of another and the sides adjacent to those angles are proportional,
C-14-Q075In the proof of Theorem 34, by —
C-14-Q076After establishing , we conclude —
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

The relation $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ in the proof follows from —

aTheorem 28 (corollary 1)

bPythagoras' theorem

cSAS congruence

dparallelogram law

ব্যাখ্যা

With $PQ \parallel BC$ , the line divides the two sides of $\triangle ABC$ proportionally, which is exactly the result of Theorem 28 (corollary 1). This yields $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ .

The relation ABAP=ACAQ\dfrac{AB}{AP} = \dfrac{AC}{AQ}APAB​=AQAC​ in the proof follows from —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The relation ABAP=ACAQ\dfrac{AB}{AP} = \dfrac{AC}{AQ}APAB​=AQAC​ in the proof follows from —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The relation $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ in the proof follows from —

The relation $\dfrac{AB}{AP} = \dfrac{AC}{AQ}$ in the proof follows from —