In triangle ABC to the right, if BC = 3 and AC = 4, then what is the

A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.

Complete solution:
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?
A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3
Important property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

So, \(\frac{CD}{AC}=\frac{AC}{BC}\) --> \(\frac{CD}{4}=\frac{4}{3}\) --> \(CD=\frac{16}{3}\).

Answer: D.

For more on this subject please check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
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Well i have a problem with similar triangles coz i sometimes make mistakes on the common sides. hence alternative approach for this problem..

Considering Triangle ACD - AC^2 + CD^2 = AD^2

Considering Triangle ABD - AB = sqrt(4^2 + 3^2) = 5 (Pythagorean triplet so you dont really have to do the math on the test)

Considering Triangle BAD - AB^2 + AD^2 = BD^2

25 + ac^2 + cd^2 = (3 + cd)^2

=> 25+16+cd^2= 9 + 6cd +cd^2

=>32 = 6cd

cd = 16/3

D- 16/3

we can use pythagoras theorem to solve this. AB we will be 5.
Let CD = x then AD = sqrt ( 16 + x^2)
in Triangle BAD we have AB^2 + AD^2 = BD^2 => 25 + 16 + x^2 = (3+x)^2
solving the above we get x= 16/3

This is a MGMAT Question,
OA is D and OE is as below. Hope it helps.

Because angles BAD and ACD are right angles, the figure above is composed of three similar right triangles: BAD, ACD and BCA. [Any time a height is dropped from the right angle vertex of a right triangle to the opposite side of that right triangle, the three triangles that result have the same 3 angle measures. This means that they are similar triangles. See below for further explanation.]

To solve for the length of side CD, we can set up a proportion, based on the relationship between the similar triangles ACD and BCA:

BC/CA = CA/CD

3/4 = 4/CD

CD = 16/3
The correct answer is D.

Addendum: Let's look at how we know that triangles ACD and BCA are similar.

1) Let's say that <CDA is x degrees, and <DAC is y degrees. Since <ACD is 90 degrees, and the sum of all the interior angles in a triangle is 180, we know that x + y = 90.

2) Now let's look at <BAC. We know that <BAC + <DAC = 90, since <BAD is labeled as a right angle. We also know that <DAC is y degrees (from step 1), and that x + y = 90. Putting these facts together, we know that <BAC is x degrees.

3) We know <ACB is a right angle, since <ACD is a right angle. Since <ACB is a right angle, <BAC + <CBA = 90. Given that <BAC is x degrees, <CBA must be y degrees.

4) To summarize, <CAB has the same measure as <CDA (x degrees) , and <CBA has the same measure as <DAC (y degrees). This means that in similar triangles CAB and CAD, side BC of CAB corresponds to side CA of CAD, and side CA of CAB corresponds to side CD of CAD.

Thus, BC/CA = CA/CD.

Again, the correct answer is D.

all three triangles abc, acd & abd are similar.
so, \(\frac{4}{3} = \frac{cd}{4}\) ... cd = \(\frac{16}{3}\)

it can't be \(\frac{4}{3} = \frac{4}{cd}\) ... cd = 3, because in that case \(5^2+5^2 = 6^2\) is not true

You can immediately tell that AB=5 because ABC is a 3-4-5 triangle. Label CD x, and AD y.

You get:

1) \(5^2+y^2=(3+x)^2\)
2) \(4^2+x^2=y^2\)

Plug in the definition of \(y^2\) from (2) into 1 and solve. You get 16/3.

Hi

I need a quick clarification on the concept of perpendicular bisector.
With a perpendicular bisector, the bisector always crosses the line segment at right angles
If any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?

So here BC should be the perpendicular bisector and the AC=CD=3 right ?

Please let me know what am missing here.

I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ?

If the 3 triangles are proportional, why can't I solve using the ratio:

\(\frac{AD}{AB} = \frac{AB}{(X+3)}\)
\(\frac{4}{5} = \frac{5}{(X+3)}\)
4(X+3) = 25
4X + 12 = 25
X = \(\frac{13}{4}\)

In similar triangles the ratio of the corresponding sides are equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

The ratios you are equating are not of corresponding sides. If you want to equate AD/AB then it should be AD/AB=AC/BC --> AD/5=4/3 --> AD=20/3. Also AD/AB=CD/AC --> (20/3)/5=CD/4 --> CD=16/3.

I merged this thread with an earlier discussion of the same question, so check this post: in-triangle-abc-if-bc-3-and-ac-4-then-what-is-the-126937.html#p1050155 it might help to clear your doubts.

Please ask if anything remains unclear.
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Hi Bunuel,

How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this can be proved thro similarity, however I wanted to understand from your concept, which you have mentioned below.
Pls explain.

Thanks
H

<B+<D+<A=180, since <A=90 then <B+<D=90;
Similarly in triangle ABC: <B+<BAC=90 since <B=90-<D then (90-<D)+<BAC=90 --> <BAC=<D.

Hope it's clear.
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Learn this property:

AC^2=BC*DC

=>DC=16/3

the given solution is neat and simple but it didnt strike me when solving
anyways here is another alternate way though i admit its lengthy calculation
let us assume CD=x
thus, AD=sqrt(x^2+16)
also in traingle ABD, BD^2=AB^2+AD^2
(x+3)^2=25+(x^2+16)
x=32/6=16/3

(D)

This part is clear: triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles)

However, how does one determine which angles are equal? Except 90 degree angles of both triangles, i could not seem to follow how exactly other angles became equal?

Both smaller triangles are similar to the large triangle. So they are similar to each other too.

In triangles BAD and BCA,
Angle BAD = BCA (90 degrees)
and angle B is common in both
So by AA, triangles BAD and BCA are similar

Similarly, in triangles BAD and ADC,
angle BAD = ACD (90 degrees)
and angle D is common in both
So by AA, triangles BAD and ACD are similar

So triangle BAD is similar to triangle BCA and ACD so triangle BCA is similar to triangle ACD too.
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Hi Karishma,

Thanks for explaining this. I understand how 3 triangles are similar to each other. However how do we determine which side is similar to which in order to set up the ratio.

for example, if we take two smaller triangles - Triangle ABC and ADC....I think side AB is corresponding to AD, BC is corresponding to CD and AC is common. Is that correct? if yes, how do I find which side is corresponding in the smaller triangle with respect to the bigger triangle?

Many thanks in advance.

There is a very simple method of finding out the corresponding sides in similar triangles.

Say, you have two triangles ABC and DEF. You find by AA that the triangles are similar. All you have to do is name the triangles the way the angles are equal.
Say angle A = angle E, angle B = angle D and and hence angle C = angle F.

Then we write:
triangle ABC is similar to triangle EDF.
Now you have the corresponding sides. That is, AB/ED = BC/DF = AC/EF

In the question above,
triangles BAD and BCA are similar
So BA/BC = AD/CA = BD/BA

triangles BAD and ACD are similar
So BA/AC = AD/CD = BD/AD

triangles BCA is similar to triangle ACD
So BC/AC = CA/CD = BA/AD
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such questions where there is a 90 degree angle in a triangle can always be solved easily by drawing a circle .
as we draw circle and extend AC to point E we will get a triangle BED , exact copy of triangle BAD --Eq1.

as shown in the attached image , Triangle ABC and Triangle ECD are similar as all angles are equal.
so \(\frac{EC}{BC} = \frac{CD}{AC} = \frac{ED}{AB}\)

3x=4y --> \(x=\frac{4}{3}*y\)
y=4 as shown in eq1
so x= \(\frac{16}{3}\)

Hi Bunuel,

please explain me the correlation between 30-60-90(1:sq(3):2), 45-45-90(1:1:sq(2)) with pythagorean triplets(3-4-5, 7-24-25). I am confused because if we have a 90deg triangle, and 2 sides 3,4 we can put other side as 5? If so then it becomes 30-60-90 triangle right? then why we are not able to correlate 1:sq(3):2 with 3:4:5?
because we need to commonly multiply the ratio, so if 1*3 then sq(3) must also multiply by 3 which is not equal to 4..

Because I had tried an another approach which gives a wrong ans... Need clarification.. If ABC is a right triangle with 3,4 then other side must be 5. So Angle BAC will be 30 deg. which makes Ang CAD as 60 and CDA as 30.. So we have 30-60-90 triangle on ACD. If thats the case, then CD must be 4*sqrt(3) right? what's wrong in my approach?

Please help..