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Standard IX

Mathematics

Intersecting and Non-Intersecting Lines

A Δ A B C is ...

Question

A ∆ABC is given. If lines are drawn through A, B, C, parallel respectively to the sides BC, CA and AB, forming ∆PQR, as shown in the adjoining figure, show that $B C = \frac{1}{2} Q R$ .

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Solution

BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA ...(i)
Similarly, BC || AR and AB || CR.
i.e., BCRA is a parallelogram.
∴ BC = AR ...(ii)
But QR = QA + AR
From (i) and (ii), we get:
QR = BC + BC
⇒ QR = 2BC
∴ BC = $\frac{1}{2} Q R$

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Similar questions

Q. In

Δ A B C

lines are drawn through

A, B

and

C

parallel to sides

B C, C A

and

A B

respectively forming a

Δ P Q R

prove that

B C = \frac{1}{2} Q R

In the adjoining figure,

A B C

is triangle. Through

A,

B

and

C

lines are drawn parallel to

B C,

C A

and

A B

respectively, which forms a

Δ P Q R .

Show that

2 (A B + B C + C A) = P Q + Q R + R P .

Q. Through A, B and C lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a

Δ A B C

as shown in figure. Show that

B C = \frac{1}{2} Q R

In the adjoining figure, $Δ A B C$ is a triangle and through A, B, C lines are drawn, parallel respectively to BC, CA and AB, intersecting at P, Q and R. Prove that the perimeter of $Δ P Q R$ is double the perimeter of $Δ A B C$ .

Q. Question 7
Through A, B and C lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a

Δ A B C

as shown in figure. Show that

B C = \frac{1}{2} Q R

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A ∆ABC is given. If lines are drawn through A, B, C, parallel respectively to the sides BC, CA and AB, forming ∆PQR, as shown in the adjoining figure, show that BC=12QR.

BC || QA and CA || QB i.e., BCQA is a parallelogram. ∴ BC = QA ...(i) Similarly, BC || AR and AB || CR. i.e., BCRA is a parallelogram. ∴ BC = AR ...(ii) But QR = QA + AR From (i) and (ii), we get: QR = BC + BC ⇒ QR = 2BC​ ∴ BC = 12QR

A ∆ABC is given. If lines are drawn through A, B, C, parallel respectively to the sides BC, CA and AB, forming ∆PQR, as shown in the adjoining figure, show that $B C = \frac{1}{2} Q R$ .

BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA ...(i)
Similarly, BC || AR and AB || CR.
i.e., BCRA is a parallelogram.
∴ BC = AR ...(ii)
But QR = QA + AR
From (i) and (ii), we get:
QR = BC + BC
⇒ QR = 2BC
∴ BC = $\frac{1}{2} Q R$