Geometry proof Mid point theorem

[mathlearn]

Hi,I have been stuck on this problem

The midpoints of the sides AB and AC of the triangle ABC are P and Q respectively. BQ produced
and the straight line through A drawn parallel to PQ meet at R. Draw a figure with this information
marked on it and prove that, area of ABCR = 8 x area of APQ.

I drew a figure with all the information included,

View attachment 5790

and I am having trouble saying that ABCR = 8 x area of APQ. & I think it has got something to do with The Midpoint theorem.Try your best to explain your answer

Many thanks:)

 

[Evgeny.Makarov]

and I am having trouble saying that ABCR = 8 x area of APQ.

You just said it! (Smile)

If we denote the middle of $BC$ by $S$, it is easy to prove that segments $PQ$, $PS$ and $QS$ divide $\triangle ABC$ is into four equal (i.e., congruent) triangles. Thus, the area of $\triangle APQ$ is $1/4$ of the area of $\triangle ABC$. Another fact that is needed is that $\triangle ABC=\triangle ARC$.

 

[mathlearn]

Many Thanks,
I see that the \(\displaystyle \triangle APQ \equiv \triangle PBS \equiv \triangle APQ \equiv \triangle QSC \) triangles are congruent to each other by (SSS)

and I updated my picture

View attachment 5791

Can you help me to say that \(\displaystyle \triangle ABC=\triangle ARC\) and to continue the proof.

I Think there are two ways of accomplishing it but not sure of them. The first one would be , ABCR is a parallelogram and \(\displaystyle \triangle ABC = \frac{1}{2} ABRC\)

and the other way not sure whether it's correct; it would be to say BQ=QR somehow.

Can you help me to proceed.

Many Thanks :)

 

[Evgeny.Makarov]

Can you help me to say that \(\displaystyle \triangle ABC=\triangle ARC\) and to continue the proof.

This should say, "Can you help me prove that \(\displaystyle \triangle ABC=\triangle ARC\)..."

I Think there are two ways of accomplishing it but not sure of them. The first one would be , ABCR is a parallelogram and \(\displaystyle \triangle ABC = \frac{1}{2} ABRC\)

ABCR is definitely a parallelogram, the question is just how to show it. It may depend on the facts that have already been proved in your course.

It is not good to say \(\displaystyle \triangle ABC = \frac{1}{2} ABRC\) because the operation of multiplying a quadrangle by a number is not defined.

and the other way not sure whether it's correct; it would be to say BQ=QR somehow.

This is also true, but requires a proof.

One way is to note that $\angle BCA=\angle CAR$ as alternate angles at parallel lines and $\angle BQC=\angle RQA$ as vertical angles. Therefore, $\triangle AQR=\triangle CQB$ by ASA (the equal sides are $AQ$ and $CQ$). In particular, $AR=BC$. Then $\triangle ABC=\triangle CRA$ by SAS (the equal angles are $\angle BCA$ and $\angle CAR$).

 

[mathlearn]

Many Thanks :) for your clear explanation,

So now \(\displaystyle \triangle ABC \equiv \triangle CRA\) Under (SAS);

Now it is obvious that ABCR = 8 x area of APQ;

How should I say that ,

Can you help me to proceed.

Many Thanks :)

 

[Evgeny.Makarov]

Now it is obvious that ABCR = 8 x area of APQ;

How should I say that

You just said that. It would be more correct to say that the area of $ABCR$ is 8 times the area of $APQ$.

I am not sure what you need help with.

 

[mathlearn]

Many Thanks :)

Now I think the problem should ended by finally writing "ABCR = 8 x area of \(\displaystyle \triangle\)PQR " after saying \(\displaystyle \triangle ABC≡ \triangle CRA under (SAS)\) & thanks for your very clear explanation on the problem.

Many Thanks:)

 

[Evgeny.Makarov]

Now I think the problem should ended by finally writing "ABCR = 8 x area of \(\displaystyle \triangle\)PQR "

Once again, the area of $ABCR$. A quadrangle can't equal a number.

 

[mathlearn]

Hi,

\(\displaystyle \therefore area of ABCR = 8 times the area of \triangle PQR \)

So that's It problem solved.

Many Thanks for your detailed and clear explanations (Smile);

 

[raj2sherma]

Proof :
ABCR = 2ABR∆ (Diagonal bisect the area of a parallelogram)
ABR∆ = 2ABQ∆ ( Base BQ is half from BR and height is same for both triangles)
ABQ = 2APQ∆ ( Base AP is half from AB and height is same for both triangles)
∴ ABCR = 2ABR∆ = 2 * 2ABQ∆
ABCR = 4ABQ∆ = 4 * 2APQ∆
ABCR = 8APQ∆