Question

in a triangle ABC point qp and r are the midpoints of the sides ab bc and ac respectively show that segment QR bisect segment AP . urgent give answer super faster​

Answer 1

Answer:

Question

P,Q and R are respectively the mid points of sides BC, CA and AB of triangle ABC. PR and BQ meet at X. CR and PQ meet at Y. prove that XY=1/4BC

Answer

Given

ABC is a Triangle.

P is the m.p of BC

Q is the m.p of CA

R is the m.p of AB

To prove

XY =$\frac{1}{4}$ BC

Proof

In ΔABC

R is the midpoint of AB.

Q is the midpoint of AC.

∴ By Midpoint Theorem,

RQ║BC

RQ║BP → 1 [Parts of Parallel lines]

RQ = $\frac{1}{2}$ BC → 2

Since P is the midpoint of BC,

RQ = BP → 3

From 1 and 3,

BPQR is a Parallelogram.

BQ and PR intersect at X

Similarly,

PCQR is a Parallelogram.

PQ and CR intersect at Y.

\implies X and Y are Midpoints of sides PR and PQ respectively.

In ΔPQR

X is the midpoint of PR

Y is the midpoint of PQ

∴ By Midpoint Theorem,

XY = $\frac{1}{2}$ RQ

From 3,

XY =$\frac{1}{2}$ +$\frac{1}{2}$ BC

XY = $\frac{1}{4}$ BC

hence proved ...

Answer 2

Answer:

P,Q and R are respectively the mid points of sides BC, CA and AB of triangle ABC. PR and BQ meet at X. CR and PQ meet at Y. prove that XY=1/4BC

Answer

Given

ABC is a Triangle.

P is the m.p of BC

Q is the m.p of CA

R is the m.p of AB

To prove

XY =\frac{1}{4}

4

1

BC

Proof

In ΔABC

R is the midpoint of AB.

Q is the midpoint of AC.

∴ By Midpoint Theorem,

RQ║BC

RQ║BP → 1 [Parts of Parallel lines]

RQ = \frac{1}{2}

2

1

BC → 2

Since P is the midpoint of BC,

RQ = BP → 3

From 1 and 3,

BPQR is a Parallelogram.

BQ and PR intersect at X

Similarly,

PCQR is a Parallelogram.

PQ and CR intersect at Y.

\implies X and Y are Midpoints of sides PR and PQ respectively.

In ΔPQR

X is the midpoint of PR

Y is the midpoint of PQ

∴ By Midpoint Theorem,

XY = \frac{1}{2}

2

1

RQ

From 3,

XY =\frac{1}{2}

2

1

+\frac{1}{2}

2

1

BC

XY = \frac{1}{4}

4

1

BC

hence proved ...