Question

Question
In the given figure ABCD is a Parallelogram in which P is the mid point of DC and Q is a point on AC such that CQ = ⅕ AC. If PQ produced meets BC at R. Then proof R is a mod point of BC.
Refer to the attachment​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question}}}$

Here the Concept of Properties of Parallelogram has been used. Here we see we are given the different points and their relation in the parallelogram. Using this we can firstly find a relation between the different sides and then prove the required things.

Let's do it !!

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★ Correct Question :-

In the given figure ABCD is a Parallelogram in which P is the mid point of DC and Q is a point on AC such that CQ = ¼ AC. If PQ produced meets BC at R. Then proof R is a mod point of BC.

Refer to the attachment

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★ Solution :-

Given,

» Mid - point of DC = P

» Mid - point of AC = Q

» There is a parallelogram ABCD whose diagonals are AC and BD

» CQ = ¼ × AC

✒ Property :: In a Parallelogram, the diagonals bisect each other.

According to this property ;

~ For Diagonal AC :

→ AO = OC

~ For Diagonal BD :

→ BO = OD

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~ For Diagonal AC ::

We already know that,

→ AO = OC

Then,

$\\\;\bf{\rightarrow\;\;\green{OC\;\;=\;\;\dfrac{1}{2}\;AC}}$

Let this be equation i)

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~ For relation between CQ and OC ::

Its given that,

$\\\;\bf{\rightarrow\;\;\red{CQ\;\;=\;\;\dfrac{1}{4}\;AC}}$

Taking ½ in common, we get

$\\\;\bf{\Longrightarrow\;\;CQ\;\;=\;\;\dfrac{1}{2}\:\bigg(\dfrac{1}{2}\;AC\bigg)}$

Applying equation i) here, we get

$\\\;\bf{\Longrightarrow\;\;CQ\;\;=\;\;\dfrac{1}{2}\:\bigg(OC\bigg)}$

$\\\;\bf{\Longrightarrow\;\;\orange{CQ\;\;=\;\;\dfrac{1}{2}\:OC}}$

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~ For the required proof ::

Let's take the ∆DOC first.

Here we see that P is midpoint of CD and Q is midpoint of OC.

✒ Mid - Point Theorem :: By this rule if a line passes through midpoints of the two opposite sides of triangle, then that line is parallel to the third line.

By this, we get,

→ PQ || DO

Let this be equation iii)

Now we know that OD = OB. This means,

→ PQ || OB

Also,

✒ Converse of Mid Point Theorem :: It states that if a line passes through the mid of a side of triangle which meets at the other opposite side of triangle and is parallel to the third side, then the point at which the line meets the opposite side is mid - point of that side.

From this, we can derive that,

→ OR || OB

Thus, R is the midpoint of side BC.

$\\\;\underline{\boxed{\bf{\mapsto\;\;\purple{R\;is\;mid\:-\:point\;of\;BC}}}}$

>> Hence proved !!

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★ More to know about Parallelogram :-

$\\\;\sf{\leadsto\;\;Area\;of\;Parallelogram\;=\;Base\;\times\;Height}$

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Parallelogram\;=\;2(Length\;+\;Breadth)}$

• Opposite Sides of parallelogram are equal.

• Sum of Adjacent Angles of parallelogram are equal.

• A parallelogram is a four sides figure.

• Diagonal of a parallelogram divides the parallelogram into two triangles of equal areas.

• Parallel Lines are the lines which have no intersection with each other and run straight to each other.

Answer 2

$\large\bf{\underline\red{Good \: Afternoon♡}}$

Question :-

In the given figure ABCD is a Parallelogram in which P is the mid point of DC and Q is a point on AC such that CQ = ¼ AC. If PQ produced meets BC at R. Then proof R is a mod point of BC.

Answer :-

We know that the diagonals of a parallelogram bisect each other.

Therefore,

CS = ½ AC ....(i)

Also, it is given that CQ = ¼ AC ...(ii)

Dividing equation (ii) by (i), we get,

$\frac{CQ}{CS} = \frac{ \frac{1}{2} AC}{\frac{1}{2} AC}$

Or, CQ = ½ CS

Hence, Q is mid point of CS.

Therefore, according to midpoint theorem in ∆CSD

PQ || DS

If PQ || DS, we can say that

QR || SB

In ∆ CSB,

Q is midpoint of CS

QR ‖ SB.

Applying converse of midpoint theorem , we conclude that R is the midpoint of CB.