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Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side ​

Answer 1

Answer:

Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third and half as long

vectors triangles

The task is to prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. (Or in vector notation PQ = AB / 2). It should be proved using some vector algebra but I am not sure how to go about doing it. A (crude) visualization:enter image description here

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Jun 8 '15 at 10:29

pseudomarvin

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Have you already studied similarity (of triangles)? – Timbuc Jun 8 '15 at 10:30

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Can you state a vector expression for point P given the vectors from the origin to points A, B, and C? (Hint: it is the average of A and B.) Can you do the same for point Q? Then get expressions for vectors PQ−→− and AB−→−. – Rory Daulton Jun 8 '15 at 10:32

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With only basic geometry:

If you've already studied similarity of triangles it is pretty easy: comparing triangles ΔCAB,ΔCPQ :

12=CPCA=CQCB,and the angle∠Cis common to both triangles

By similarity theorem , ΔCAB∼ΔCPQ , and thus

PQAB=12⟺2PQ=AB

That PQ||AB follows from the fact that similar triangles have the same angles, and thus ∠CAB=∠CPQ .

With vectors:

Put u:=CA→,v:=CB→ , then we get:

CP→=12u,CQ→=12CB,AB→=−u+v=−(v−u)

so

PQ→=−12+12b=−12(u−v)=12AB→

and we're done as the last line both proves the middle segment is parallel to AB and its length is half that of the latter.

Step-by-step explanation:

follow me dear

Answer 2

Answer:

Given=in triangle ABC

AP =BP AND AQ=CQ

TO PROVE=PQ PALLEL BC AND OQ HALF OF BC

CONSTRUCTION= PRIDUCEC PQ TO R SUCH THAT PQ =QR ABD JOIN CR

PROOF= IN TRIANGLE AQP AND TRIANGLE CQR

AQ=CQ(GIVEN)

PQ=QR(GIVEN)

ANGKE1=ANGLE2(V.O.A)

TRI. AQP IS CONGRUNENT TO TRI.CQA(S.A.S)

ER=AP(C.P.C.T)

ANGLE3=ANGLE4(C.P.C.T)

BUT THESE ARE ALT.

CR PARLEL AP

CR PALLEL AB

CR PALLEL BP-1

CR PALLEL AP

BUT AP=BD( GIVEN)

CR=AP-2

FROM 1 AND 2

PRCB IS A PARALOGRAM

PR=BC AND PR PALLEL BC

PQ+ QR=BC

PQ+PQ=BC

2 PQ=BC

PQ=1/2 BC

HOPE IT WILL HELP YOU