The problem is as follows:

The figure shows a square $ABCD$ which has an area of $120\,cm^2$. Assuming that you intend to find the area of the triangle $\triangle{APQ}$.

Sketch of the problem

Which of the following information is enough to solve what it is being asked?.

I. $M$ is midpoint of the segment $AD$.

II. $N$ is midpoint of the segment $CD$.

Therefore in order to get an answer for the problem. Select one of the choices.

$\begin{array}{ll} 1.&\textrm{Information I is enough to solve this problem}\\ 2.&\textrm{Information II is enough to solve this problem}\\ 3.&\textrm{Each information, either I or II by separate is enough to solve this problem}\\ 4.&\textrm{In order to solve the problem it is needed both information I and II}\\ \end{array}$

I'm not sure exactly how to solve this problem or which choice would be the right one. To my understanding it is needed more information to solve this problem. I really don't know exactly what is needed to get the area for that triangle. Can someone help me here?. I feel confused.

The thing is that the only thing I was able to get is that the edge of this square is $2\sqrt{30}$ centimeters. But I don't know if this can be used as it is to get the area of that little triangle.

Since this problem is intended to be solved relying only in euclidean geometry. I am requesting assistance for such approach.

Thus can someone help me here please?. Plese since, I am not good with euclidean geometry it would help me a lot that an answer would include the steps with some sort of commentary so I can understand why it was used such conclusion. Thus all and all, can someone help me?.

  • $\begingroup$ Please work with similar triangles. $\endgroup$ – Math Lover Apr 8 at 3:23
  • 2
    $\begingroup$ Suppose that (I) is enough. Without (II), N will be allowed to run freely along CD. This means there is no definite answer for the area of $\triangle APQ$. $\endgroup$ – Mick Apr 8 at 3:30
  • $\begingroup$ Look at triangles $ \ ABC \ $ and $ \ ADN \ $ ; that will tell you what angle $ \ PAQ \ $ is. $ \ ADN \ $ is congruent to another triangle which will help you find that $ \ AQP \ $ is a right angle. Is there anything that tells you about one side of triangle $ \ APQ \ $ ? Could you say any of that for sure if both I and II weren't known? $\endgroup$ – boojum Apr 8 at 3:54
  • $\begingroup$ Construct the figure with GeoGebra (free) and see what happens if $M$ and $N$ are moved. $\endgroup$ – Intelligenti pauca Apr 8 at 7:19
  • 2
    $\begingroup$ If the goal is to "find the area of $\triangle APQ$", then neither midpoint statement is necessary; the area can be expressed symbolically in terms of, say, $|AM|$ and $|DN|$. Even if the goal is to show that $\triangle APQ$ has a particular area, then midpoint-ness per se still isn't necessary: the area when $M$ and $N$ are midpoints can also be achieved when $M$ and $N$ are in other locations. ... This seems a poorly-posed question to me. $\endgroup$ – Blue Apr 8 at 11:12

As I noted in a comment, it's not necessary to know that $M$ and $N$ are midpoints to find the area of $\triangle APQ$; nor is the midpoint-ness of $M$ and $N$ required to attain a particular area. So, I consider the question poorly-posed. Nevertheless, to the extent that the exercise is suggesting that we know midpoint-ness or nothing about $M$ and $N$, then: Both conditions I and II are required.

For the sake of near-completeness, I'll note that if we have simply that $|AM|=\mu |AB|$ and $|DN|=\nu |AB|$ for some $\mu$ and $\nu$, then one can determine (without too much trouble, but more than is worth typing-up here) that $$\frac{|\triangle APQ|}{|\square ABCD|}=\frac{\mu^2(1-\nu)}{2(1+\mu)(1+\mu\nu)}$$ Here we can see that the ratio depends upon both $\mu$ and $\nu$ (and thus the locations of both $M$ and $N$); also, a given ratio can be achieved from different $\mu$ and $\nu$ values. In the case of midpoints, where $\mu=\nu=1/2$, this ratio reduces to $1/30$, consistent with other answers.

That said, once we've decided that $M$ and $N$ are midpoints, there's a diagrammatic derivation of the result:

enter image description here

$$|\square ABCD| = 5\;|\square QRST| = 5\cdot 6\;|\triangle APQ| = 30\:|\triangle APQ|$$

(Note: Since congruent triangles $\triangle ABM$ and $\triangle DAN$ are rotated $90^\circ$ from one another, so are their respective hypotenuses; this is how we know $\square QRST$, and the rest, are squares. Also, a simple similarity argument guarantees that $P$ and $U$ are trisection points of $\overline{QR}$ and $\overline{ST}$.)

  • 1
    $\begingroup$ I love the graphical approach here. What did you use for making that? Geogebra? Also glad to see another answer not requiring trig. $\endgroup$ – Eric Snyder Apr 10 at 22:31
  • $\begingroup$ @EricSnyder: Thanks. Graphical approaches are kinda my thing. :) The diagram was indeed made with GeoGebra. $\endgroup$ – Blue Apr 10 at 23:27

I will illustrate your question with help of square of side length $6 units$ placed in first quadrant as shown below, lets see all the cases one by one.

CASE 1 : $M$ and $N$ are mid-points.

enter image description here

We see that in this case that you can always find the equation of lines hence co-ordinates of triangles therefore the area moreover I can assure you line $BM$ and $AN$ will be perpendicular so we have right angled triangle case hence you can easily find out the area .

CASE 2 : $M$ is mid-point.

Here you can look in above diagram that you can be assure of co-ordinate location of $P$ and point $Q$ can slide so angle $APQ$ is constant and your side length $AP$ with these information you can never find the area with any formula.

CASE 3 : $N$ is mid-point.

Here you can look in above diagram that as $M$ will slide both $P$ and $Q$ will change so you don't have sufficient information to remark or validate the area.

So we need both information in this question.

Now how we can find area of that triangle, so for that imagine your case in first quadrant plot the things in exactly same way as I did and change the side length to $2\sqrt{30}$ since area is $120 unit^2$ now you can find out equation of lines $BM$, $AN$, $AC$ hence you can find out intersection points $P$ and $Q$ after that for right angled triangle $APQ$ you can find out its height and base and hence required area.

The exact area will be $4unit^2$ when you will try it on your own with the method that I told above however you can cross check the intermediates that I am listing down for your help.

Equations of lines:

AC: $y=x$

AN: $y=x/2$

BM: $y=-2(x-\sqrt{30})$

Point of Intersections:

A: $(0,0)$

P: $(2(\sqrt{30})/3,2(\sqrt{30})/3)$

Q: $(4(\sqrt{30})/5,4(\sqrt{30})/10)$

Side Lengths:

PQ: $\sqrt{8/3}$

AQ: $\sqrt{24}$

AREA= $(1/2)*\sqrt{8/3}*\sqrt{24}$ = $4unit^2$.

  • $\begingroup$ I think with some effort you can conclude using congruence that $\angle PQA=90^\circ$ but in order to find the area of such tiny triangle. You need to know the lengths of the segments $PQ$ and $AQ$, how do you know these segments?. Can you help me here please. $\endgroup$ – Chris Steinbeck Bell Apr 8 at 21:02
  • $\begingroup$ You can refer last paragraph I have updated it, hopefully I think now your query would have been resolved. $\endgroup$ – solver Apr 9 at 7:37

Why do we need information about both M and N?

P is free to slide along AC until we lock $\angle ABP$ or $\angle ABM$. Similarly, Q is free to slide along BM until we lock $\angle DAQ$ or $\angle DAN$. Therefore, choice (4) is correct.

How to find $S_{\triangle APQ}$?

We will prove $\triangle APQ$ is a right triangle and find the lengths of its two legs.

Set AB=2 cm for easy calculation first and scale back to$\sqrt {120}$ cm later..


$\triangle BAM \cong \triangle ADN (SSS)$ $\triangle BAM \sim\triangle AQM (AAA), AQ \perp BM, \frac{QM}{AM}=\frac{AM}{BM} $ \begin{align}QM=\frac{1}{\sqrt 5},\frac{AQ}{QM}=\frac{AB}{AM},\color {blue}{AQ}=2QM=\frac{2}{\sqrt 5}\end{align}


  1. $\triangle APO \sim \triangle BPO (\theta=\theta'=\theta'' ), \,\frac{AP}{AQ}=\frac{BP}{BO}, \frac{AP}{(\frac{2}{\sqrt 5})}=\frac{BP}{\sqrt 2} $

  2. $\triangle APQ$ with PT $AP^2-(\frac{2}{\sqrt 5})^2=BQ^2$

  3. $BM=BP+PQ+QM= AP+PQ+\frac{2}{\sqrt 5}=\sqrt 5$

$\color {blue} {PQ}=\frac{1}{3}AQ=\frac{2}{3\sqrt 5}\\$ $S'_{\triangle APQ}=\frac{1}{2}\color{blue}{PQ\times AQ}=\frac{2}{15}$

$\to S_{\triangle APQ}= S'_{\triangle APQ} \times \frac{120}{4}=\frac{2}{15} \times \frac{120}{4}=\boxed {4 cm^2}$

enter image description here

  • $\begingroup$ Okay I understood that If we do not lock those positions we cannot find the area. But once we declare that those two statments are needed. How can we do check and find that area?. Others have noted that $\angle PQA=90^\circ$ but how do we prove that? can you please explain these doubts which I have?. I am still confused on how to get that area. $\endgroup$ – Chris Steinbeck Bell Apr 8 at 20:56

A general approach to take for a question such as this:

We are given a problem to solve, with sets of statements 1, 2 and 3. In our example, Statement Set 1 is that M is the midpoint of AD, Statement Set 2 is that N is the midpoint of CD, and Statement Set 3 is all the other information (such as that ABCD form a square, that the area of square ABCD is 120, that M lies on AD, that N lies on CD, that P is the intersection of AC and BM, that Q is the intersection of AN and BM, and that APQ aren't colinear.

First, check to confirm that 1,2, and 3 are sufficient to solve for the area. To do this, we can simply declare that A is the origin, that B lies on the positive part of the Y axis, and that D lies on the positive part of the X axis. We can then use the statements to find the exact coordinates of B, C, D, M, N, P, and Q, and can use Gauss's shoelace formula to find the area of APQ. Next, create alternate sets of information, 1a and 2a. For example, let 1a state that M is X times as close to D as to A, and let 2a state that N is Y times as close to D as to C. Then, try to solve our system given the information contained from 1a,2,3. Is it still solvable (in our case, yes, as we can use a similar method to locate all the points) in terms of X. If it is not solvable, or if it is solvable but dependent on the value of X, than 1 is necessary to solve the problem. Otherwise, given 2, 1 is not necessary. Similarly, we can test 1,2a,3 to check if 2 is necessary when given 1. If and only if we find that neither is necessary when given the other, we then test 1a,2a,3 to see if either 1 or 2 is necessary

In our particular instance, it is trivial to show that if we choose a point M' between A and M (and create the points P' and Q'), then AP'Q' can be found within APQ, but can be made arbitrarily small. This means that given 2, 1 remains necessary. We can then similarly show that by placing N' between N and C, creating Q", that APQ" is entirely within APQ and can be made arbitrarily small, meaning that even given 1, 2 remains necessary. As such, both are necessary


enter image description here

In this approach, we won't need much in the way of numerical values until we get close to the end; we will make use of some trigonometry, including identities. We'll call $ \ a \ $ the length of a side of the square $ \ ABCD \ \ , \ $ knowing that $ \ a^2 = 120 \ $ . The length of side $ \ AM \ $ is $ \ \frac{a}{2} \ $ from statement I ; so for angle $ \ \angle ABM \ , $ we have $ \ \tan \Theta \ = \ \frac12 \ $ (we won't care what the numerical measure of this angle is). $ \ \Delta MAB \ $ is a right triangle, so $ \ \angle AMB \ $ has measure $ \ 90º - \Theta \ . $ Since $ \ \Delta NDA \ $ is congruent to $ \ \Delta MAB \ $ (which requires statement II) , then $ \ \angle DAN \ $ also has measure $ \ \Theta \ . $ Consequently, $ \ \Delta MQA \ $ is similar to $ \ \Delta MAB \ $ , so it is also a right triangle: thus, $ \ \angle MQA \ $ and $ \ \angle AQP \ $ are right angles, and $ \ \Delta AQP \ $ is a right triangle.

Now $ \ \Delta ABC \ $ is a right isosceles triangle, so $ \ \angle CAB \ $ has measure $ \ 45º \ , $ leaving $ \ \angle QAP \ $ with measure $ \ 90º - 45º - \Theta \ = \ 45º - \Theta \ . $ We can now apply trigonometric values. The length of $ \overline{AQ} \ , $ which is the "base" of $ \ \Delta MAQ \ $ is $ \ \frac{a}{2} · \cos \Theta \ , $ which is also the "base" $ \ b \ $ of $ \ \Delta QAP \ . $ The "height" of $ \ \Delta QAP \ $ is then $ \ h \ = \ b · \tan(45º - \Theta) \ . $

At last, we obtain the area of $ \ \Delta QAP \ $ as $$ A \ = \ \frac12 · b · h \ \ = \ \ \frac12 · b · b · \tan(45º - \Theta) \ \ = \ \ \frac12 · \left[\frac{a}{2} \right]^2 · \cos^2 \Theta \ · \tan(45º - \Theta) $$

[now we apply some trig identities]

$$ = \ \frac{a^2}{8} · \frac{1}{\sec^2 \Theta} \ · \frac{\tan 45º \ - \ \tan \Theta}{1 \ + \ \tan 45º · \tan \Theta} \ \ = \ \ \frac{a^2}{8} · \frac{1}{1 \ + \ \tan^2 \Theta} \ · \frac{\tan 45º \ - \ \tan \Theta}{1 \ + \ \tan 45º · \tan \Theta} $$

$$ = \ \ \frac{a^2}{8} · \frac{1}{1 \ + \ \frac14} \ · \frac{1 \ - \ \frac12}{1 \ + \ 1 · \frac12} \ \ = \ \ \frac{a^2}{8} · \frac{1}{\frac54} \ · \frac{\frac12}{\frac32} \ \ = \ \ \frac{a^2}{8} · \frac{4}{5} \ · \frac{1}{3} \ \ = \ \ \frac{a^2}{30} \ \ , $$

which, as $ \ a^2 \ = \ 120 \ \text{cm.}^2 \ \ , $ tells us that the area of $ \ \Delta QAP \ $ is $ \ 4 \ \text{cm.}^2 \ \ . $


OK, it seems you've understood why the positions of M and N are both needed--as someone noted above, technically they wouldn't have to be midpoints, but midpoints are good enough.

As to how to solve, you seem confused about two things: (A) Why $\overline{BM}$ and $\overline{AN}$ are perpendicular, and (B) how to find the area of $\triangle APQ$, knowing (A).

That said, I'm avoiding anything other than standard geometry and simple algebra. This problem is relatively hard (for high school geometry), but it can be done without resorting to trig, without graphing the lines and using the line equations, and in fact without even calculating the lengths of the various segments.

I'm skipping a step or two here and there, but nothing essential. I'm sure Miss Studenka, my HS geometry teacher, would disapprove, but alas.

Proof that $\overline{BM} \perp \overline{AN}$


  1. $ABCD$ is a square
  2. $N$ is the midpoint of $\overline{AD}$
  3. $M$ is the midpoint of $\overline{CD}$

We can show:

  1. $AB = AD$ from the definition of a square.
  2. $\angle BAD = \angle ADC = 90°$, also from the definition of a square.
  3. $AM = DN$, from (4) and the defined midpoints.
  4. $\triangle ABM \cong \triangle DAN$, by the side-angle-side theorem.
  5. $\angle MBA = \angle NAD$, by congruence.
  6. $\angle NAB = 90° - \angle NAD = 90° - \angle MBA$, from (5) and (8)
  7. $\angle NAB + \angle MBA = 90°$, rearranging (9)
  8. $\angle NAB + \angle MBA + \angle BQA = 180°$, from the definition of a triangle.
  9. $\angle BQA = 90°$ by subtraction. $\square$

OK, that's done. How do we find the area now? First, note that $\triangle APQ$ is a right triangle, so if we know any two of its sides we can find the area. Second, we want to note that $\triangle MAB \sim \triangle MQA \sim \triangle AQB$. Let's actually show that, using the triple-angle theorem:

  1. $\angle MAB = \angle MQA = \angle AQB = 90°$ as we've already shown they're right triangles.
  2. $\angle BMA = \angle AMQ = \angle BAQ$. The first equality is true because $M,Q,$ and $B$ are collinear; the second equality we showed in (8).
  3. $\angle ABM = \angle QBA = \angle QAM$. These are true by the same logic as in (14).

Therefore all three triangles are similar. Now, for simplicity, we're going to ignore everything but A, B, Q, M, and the segments between those four points for a moment, and use the following substitutions: $AM = a, AB = b, BM = c, BQ = d, QM = e$, and $AQ = h$. Note that $h$ here is the altitude of the right triangle. Also note that $c = d+e$. Now because our triangles are similar, we know:

$$\frac{e}{h} = \frac{h}{d} \implies h^2 = de$$

$$\frac{a}{b} = \frac{h}{d} \implies \frac{a^2}{b^2} = \frac{h^2}{d^2} \implies \frac{a^2}{b^2} = \frac{de}{d^2} = \frac{e}{d} \implies a^2d = b^2e$$

But because $c = d+e$, this means

$$a^2d = b^2(c-d) \\ a^2d = b^2c - b^2d \\ (a^2 + b^2)d = b^2c \\ c^2d = b^2c \\ d = \frac{b^2}{c}$$

Similar logic but substituting the other direction gets us $e = \frac{a^2}{c}$. Now, using the side length of the square $AB = s$:

  1. $AQ = h = \sqrt{de} = \sqrt{\frac{a^2b^2}{c^2}} = \frac{ab}{\sqrt{a^2+b^2}} = \frac{\frac{s}{2} \cdot s}{\sqrt{(\frac{s}{2})^2+s^2}} = s\sqrt{\frac15}$

Now we need another side. $AP$ is easier because once again we have two similar triangles: $\triangle PAM$ and $\triangle PCB$:

  1. $\angle PAM = \angle PCB$ as opposite interior angles of parallel lines
  2. $\angle PMA = \angle PBC$ for the same reason
  3. $\angle APM = \angle BPC$ as opposite angles
  4. $\triangle PAM \sim \triangle PCB$
  5. $BC = 2 \cdot AM$ because of the midpoint
  6. $CP = 2 \cdot AP$ because the triangles are similar
  7. $3 \cdot AP = AC$ and so $AP = \frac13 \cdot AC$ by algebra
  8. $AP = \frac13 \cdot AC = \frac13 \sqrt{2s^2} = s\sqrt{\frac29}$

We now have one leg ($AQ = s\sqrt{\frac15}$) and the hypotenuse ($AP = s\sqrt{\frac29}$) of $\triangle APQ$. That's enough to get the other leg and the area: $PQ = \sqrt{AP^2 - AQ^2} = s\sqrt{\frac{1}{45}}$ making the area

$$A = \frac12 ab = \frac12 \cdot s \sqrt{\frac15} \cdot s \sqrt{\frac{1}{45}} = \frac{s^2}{30}$$

And since the problem gave us $s^2 = 120 \text{ cm}^2$, the area we want is $4 \text{ cm}^2$.

The keys to solving this without any trig are showing that the intersection at $Q$ is perpendicular, then recognizing the sets of similar triangles. (I'm shocked no other answer pointed out $\triangle BPC \sim \triangle APM$. That's what lets us skip the trig.) Of note: the calculations I showed for finding $h$ apply to all right triangles. One interesting fact I couldn't shoehorn in is that you can find the length of that altitude with the "Inverse Pythagorean Theorem":

$$\frac{1}{h^2} = \frac{1}{a^2} + \frac{1}{b^2}$$

It's in there but kind of hidden, at the point where we found $h$ from $\frac{ab}{c}$.

If you have any questions or steps you can't follow, post a comment.

  • 1
    $\begingroup$ +1 for working through a (nearly) pure Euclidean proof; following through it carefully, however, makes me thank Hipparchus and Descartes once again... $\endgroup$ – boojum Apr 11 at 2:20
  • $\begingroup$ @boojum Some small piece of my heart really did want to write the whole thing out like my geometry teacher would have wanted me to, with the full Euclidean... but it would have been so long on the page. It's already long enough! $\endgroup$ – Eric Snyder Apr 11 at 7:04

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