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

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]
26 Feb 2012, 11:07

Expert's post

1

This post was BOOKMARKED

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 [ 4.22 KiB | Viewed 14257 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 12216 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]
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}.

Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink]
17 May 2012, 04:59

1

This post received KUDOS

Expert's post

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 [ 4.22 KiB | Viewed 11937 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]
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]
20 Jun 2013, 06:18

1

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 _________________

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