Question

10th ncert chapter -2 triangles

Q8. Using Theorem 2.2, prove that the line joining the
mid-points of any two sides of a triangle is parallel
to the third side. (Recall that you have done it in
Class IX).

Theorem-2 :- If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

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Answer 1

Step-by-step explanation:

IN TRIANGLE ABC. DE line joining the triangles mid POINT.

given÷a line joining the mid points of side of a triangle.then

to prove÷the line is parallel to the third side.

proof÷D IS MID POINT (given)

AD/DB=1 ----- EQUATION (1)

AS E IS MID POINT

AE/EC=1------EQUATION( 2)

FROM EQUATION( 1 )AND( 2 )WE GET ,

AD/DB=AC/EC-----EQUATION (3)

NOW,

IN THE EQUATION (3)

RATIOS ARE equal then by B.P.T THEOREM WE SAY THAT DE PARALLEL TO BC

HENCE PROVED

Answer 2

• Proving that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

See figure 1 in the attached image.

In ΔPQR,

N is the midpoint of side PQ

And M is the midpoint of side PR

We get a line NM.

To be proved:

⇒ NM ║QR

PROOF -

In ΔPQR,

N is the midpoint of PQ, we get,

⇒ PN = QN

⇒ $\boxed{\frac{PN}{QN}=1}$   ...............(i)

Also,

M is the midpoint of PR, we get,

⇒ PM = RM

⇒ $\boxed{\frac{PM}{RM}=1}$   ...............(ii)

Now,

From the above equations (i) and (ii), we get,

$\boxed{\boxed{\frac{PN}{QN}=\frac{PM}{RM}}}$

• If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Hence,

NM ║QR (Proved)

- - -

• If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. (Proof)

See figure 2.

Given

In ΔABC,

DE is a line that divides the triangle in the same ratio

$\implies \boxed{\frac{AD}{BD}=\frac{AE}{CE}}$

To Proof:

DE ║ BC

Constriction -

Draw EN ⊥ AD and DM ⊥AB

Join BE and CD

PROOF:-

We know the area of triangle = $\boxed{\red{\frac{1}{2}\times base \times height}}$

$\implies ar(\triangle ADE) = \frac{1}{2} \times AD \times EN$

$\implies ar(\triangle BDE) = \frac{1}{2} \times BD \times EN$

$\implies \frac{AD}{BD} = \frac{\frac{1}{2}\times AD \times EN}{\frac{1}{2}\times BD \times EN} = \frac{ar(\triangle ADE)}{ar(\triangle BDE)}$

Similarly,

$\implies \frac{AE}{CE} =\frac{ar(\triangle ADE)}{ar(\triangle CDE)}$

Now,

According to the question,

$\frac{AE}{CE}=\frac{AD}{BD} \implies \boxed{ \frac{ar(\triangle ADE)}{ar(\triangle BDE)}=\frac{ar(\triangle ADE)}{ar(\triangle DEC)}}$

$\implies \frac{\cancel{ar(\triangle ADE)}}{\cancel{\bf ar(\triangle ADE)}}=\frac{\bf ar(\triangle BDE)}{ar(\triangle DEC)}$

$\implies 1 = \frac{ar(\triangle BDE)}{ar(\triangle DEC)}$

$\implies ar(\triangle DEc) = ar(\triangle BDE)$

So, the area of Δ DEC = area of Δ BDE and they have the same base.

So, They lie between parallels.

[ ∵ Triangles having the same base and lie between two parallel side have equal area.]

Hence,

DE║BC (Proved)