Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

Show Tags

26 Feb 2012, 08:42

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 ?

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 ?

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

Attachment:

splittingtriangle.jpg [ 4.22 KiB | Viewed 64598 times ]

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

splittingtriangle.jpg [ 4.22 KiB | Viewed 59116 times ]

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.

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

Show Tags

17 May 2012, 04:46

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

Bunuel wrote:

rvinodhini wrote:

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 ?

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

Attachment:

splittingtriangle.jpg

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

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

Attachment:

splittingtriangle.jpg [ 4.22 KiB | Viewed 58829 times ]

<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.

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

Show Tags

17 May 2012, 06:36

Thanks for the solution. I just wanted to know whether it requires any calculation or is it just a corollary of the property described above. Its clear now.

Bunuel wrote:

imhimanshu wrote:

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

Attachment:

splittingtriangle.jpg

<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.

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

Show Tags

20 Jun 2013, 06:18

2

This post received KUDOS

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
_________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold

Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013

GMAT - 770, Q50, V44, Oct 7th, 2013 My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

Show Tags

09 Mar 2015, 02:01

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

Show Tags

11 Mar 2015, 23:43

1

This post received KUDOS

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

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

Show Tags

24 Mar 2015, 16:13

Bunuel wrote:

rvinodhini wrote:

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 ?

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

Attachment:

splittingtriangle.jpg

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

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?

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 ?

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

Attachment:

splittingtriangle.jpg

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

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.
_________________

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]

Show Tags

27 Mar 2015, 09:54

VeritasPrepKarishma wrote:

Bunuel wrote:

rvinodhini wrote:

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 ?

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

Attachment:

splittingtriangle.jpg

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

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.

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?

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
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...