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Using statements 1 and 2, we know that AC is the perpendicular bisector of BD. This means that triangle BAD is an isosceles triangle so side AB must have a length of 5 (the same length as side AD). We also know that angle BAD is a right angle, so side BD is the hypotenuse of right isosceles triangle BAD. If each leg of the triangle is 5, the hypotenuse (using the Pythagorean theorem) must be 5 underroot 2.

If angle BAD is a right angle, what is the length of side BD?

(1) AC is perpendicular to BD

(2) BC = CD

Official answer explanation:

Using statements 1 and 2, we know that AC is the perpendicular bisector of BD. This means that triangle BAD is an isosceles triangle so side AB must have a length of 5 (the same length as side AD). We also know that angle BAD is a right angle, so side BD is the hypotenuse of right isosceles triangle BAD. If each leg of the triangle is 5, the hypotenuse (using the Pythagorean theorem) must be 5 underroot 2.

Can someone kindly explainthe underlined portion?

For such kind of graphic questions you MUST post the image. Next, please also do check the OA's when posting a question, OA for this one is C, not E.

Original question is below:

If angle BAD is a right angle, what is the length of side BD?

(1) AC is perpendicular to BD (2) BC = CD

Now, obviously each statement alone is not sufficient.

When taken together we have that AC is a perpendicular bisector. Now, if a line from a vertex to the opposite side is both perpendicular to it and bisects it then this side is a base of an isosceles triangle (or in other words if a bisector and perpendicular coincide then we have an isosceles triangle). You can check this yourself: in triangles ACD and ACB two sides are equal (AC=AC and BC=CD) and included angle between these sides are also equal (<ACD=<ACB=90) so we have Side-Angle-Side case (SAS), which means that ACD and ACB are congruent triangles, so AB=AD --> ABD is an isosceles triangle.

Next, as ABD is an isosceles triangle then AB=AD=5 and hypotenuse \(BD=5\sqrt{2}\).

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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19 Oct 2012, 12:24

If AC is perpendicular to BD, then AC is the altitude from the right angle, isn't it? That would imply that triangles ABC and ACD are similar, so that angle CDA is equal to angle ABC, thus making triangle ABD an isoeleces right triangle and you could calculate BD then. What is wrong with that argumentation?

If AC is perpendicular to BD, then AC is the altitude from the right angle, isn't it? That would imply that triangles ABC and ACD are similar, so that angle CDA is equal to angle ABC, thus making triangle ABD an isoeleces right triangle and you could calculate BD then. What is wrong with that argumentation?

Yes, the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. But the corresponding (equal) angles would be BAC and ADC (also DAC and ABC) not CDA and ABC.

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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21 Jun 2013, 07:45

Bunuel wrote:

BN1989 wrote:

If AC is perpendicular to BD, then AC is the altitude from the right angle, isn't it? That would imply that triangles ABC and ACD are similar, so that angle CDA is equal to angle ABC, thus making triangle ABD an isoeleces right triangle and you could calculate BD then. What is wrong with that argumentation?

Yes, the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. But the corresponding (equal) angles would be BAC and ADC (also DAC and ABC) not CDA and ABC.

Hope it's clear.

i dint understood these part bunuel how to find corresponding angles of similar triangles?

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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22 Jun 2013, 05:45

WarriorGmat wrote:

Bunuel wrote:

BN1989 wrote:

If AC is perpendicular to BD, then AC is the altitude from the right angle, isn't it? That would imply that triangles ABC and ACD are similar, so that angle CDA is equal to angle ABC, thus making triangle ABD an isoeleces right triangle and you could calculate BD then. What is wrong with that argumentation?

Yes, the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. But the corresponding (equal) angles would be BAC and ADC (also DAC and ABC) not CDA and ABC.

Hope it's clear.

i dint understood these part bunuel how to find corresponding angles of similar triangles?

If AC is perpendicular to BD, then AC is the altitude from the right angle, isn't it? That would imply that triangles ABC and ACD are similar, so that angle CDA is equal to angle ABC, thus making triangle ABD an isoeleces right triangle and you could calculate BD then. What is wrong with that argumentation?

Yes, the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. But the corresponding (equal) angles would be BAC and ADC (also DAC and ABC) not CDA and ABC.

Hope it's clear.

i dint understood these part bunuel how to find corresponding angles of similar triangles?

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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21 Oct 2013, 08:20

Bunuel wrote:

BN1989 wrote:

If AC is perpendicular to BD, then AC is the altitude from the right angle, isn't it? That would imply that triangles ABC and ACD are similar, so that angle CDA is equal to angle ABC, thus making triangle ABD an isoeleces right triangle and you could calculate BD then. What is wrong with that argumentation?

Yes, the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. But the corresponding (equal) angles would be BAC and ADC (also DAC and ABC) not CDA and ABC.

Hope it's clear.

Here's what I did, is my logic correct?

(1) AC is perpendicular to BD - you don't know the relationship between bc & cd though - insuff

(2) BC = CD - doesn't tell you anything, because you can't calculate what either length is.

combined: ACD is a 45-45-90, we know the hypotaneuse, and since the sides of a 45-45-90 are in the ratio of 1:1:\(\sqrt{2}\), we can solve for cd, and thus bc. So in this case, with the hypotaneuse being 5, ac & cd are both 5/\(\sqrt{2}\), and since bc=cd, we can solve, thus suff.

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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11 Dec 2013, 09:16

If angle BAD is a right angle, what is the length of side BD?

I had originally said B, thinking that because we knew CA was a bisector and that angle A was 90 degrees, there would only be one possible way to configure this triangle. Obviously, this is incorrect. You could have an isosceles triangle or you could have a triangle with a low, long line BA and a relatively short and steep line AD with B as a 90 degree angle.

(1) AC is perpendicular to BD

This tells us that AC is the altitude but little more. Insufficient

(2) BC = CD

This tells us that AC is the bisector. Triangle BAC and DAC have the same area but they could have different exterior measures for BA and DA. Insufficient.

1+2) A line drawn from a vertex to the opposite line that is both perpendicular to that line and bisects it must create a isosceles triangle with that line as the base. This means, BA = AD = 5. Furthermore, BC = CD and they can have exactly one measure...Sufficient.

If angle BAD is a right angle, what is the length of side [#permalink]

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31 Jul 2014, 09:21

Bunuel wrote:

chiragatara wrote:

If angle BAD is a right angle, what is the length of side BD?

(1) AC is perpendicular to BD

(2) BC = CD

Official answer explanation:

Using statements 1 and 2, we know that AC is the perpendicular bisector of BD. This means that triangle BAD is an isosceles triangle so side AB must have a length of 5 (the same length as side AD). We also know that angle BAD is a right angle, so side BD is the hypotenuse of right isosceles triangle BAD. If each leg of the triangle is 5, the hypotenuse (using the Pythagorean theorem) must be 5 underroot 2.

Can someone kindly explainthe underlined portion?

For such kind of graphic questions you MUST post the image. Next, please also do check the OA's when posting a question, OA for this one is C, not E.

Original question is below:

If angle BAD is a right angle, what is the length of side BD?

(1) AC is perpendicular to BD (2) BC = CD

Now, obviously each statement alone is not sufficient.

When taken together we have that AC is a perpendicular bisector. Now, if a line from a vertex to the opposite side is both perpendicular to it and bisects it then this side is a base of an isosceles triangle (or in other words if a bisector and perpendicular coincide then we have an isosceles triangle). You can check this yourself: in triangles ACD and ACB two sides are equal (AC=AC and BC=CD) and included angle between these sides are also equal (<ACD=<ACB=90) so we have Side-Angle-Side case (SAS), which means that ACD and ACB are congruent triangles, so AB=AD --> ABD is an isosceles triangle.

Next, as ABD is an isosceles triangle then AB=AD=5 and hypotenuse \(BD=5\sqrt{2}\).

on above link you have mentioned perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

what you mean by perpendicular to the hypotenuse

and in this question you are saying that it is an isoceles triangle. by using st1 then why this is not sufficient.

If angle BAD is a right angle, what is the length of side BD?

(1) AC is perpendicular to BD

(2) BC = CD

Official answer explanation:

Using statements 1 and 2, we know that AC is the perpendicular bisector of BD. This means that triangle BAD is an isosceles triangle so side AB must have a length of 5 (the same length as side AD). We also know that angle BAD is a right angle, so side BD is the hypotenuse of right isosceles triangle BAD. If each leg of the triangle is 5, the hypotenuse (using the Pythagorean theorem) must be 5 underroot 2.

Can someone kindly explainthe underlined portion?

For such kind of graphic questions you MUST post the image. Next, please also do check the OA's when posting a question, OA for this one is C, not E.

Original question is below:

If angle BAD is a right angle, what is the length of side BD?

(1) AC is perpendicular to BD (2) BC = CD

Now, obviously each statement alone is not sufficient.

When taken together we have that AC is a perpendicular bisector. Now, if a line from a vertex to the opposite side is both perpendicular to it and bisects it then this side is a base of an isosceles triangle (or in other words if a bisector and perpendicular coincide then we have an isosceles triangle). You can check this yourself: in triangles ACD and ACB two sides are equal (AC=AC and BC=CD) and included angle between these sides are also equal (<ACD=<ACB=90) so we have Side-Angle-Side case (SAS), which means that ACD and ACB are congruent triangles, so AB=AD --> ABD is an isosceles triangle.

Next, as ABD is an isosceles triangle then AB=AD=5 and hypotenuse \(BD=5\sqrt{2}\).

on above link you have mentioned perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

what you mean by perpendicular to the hypotenuse

and in this question you are saying that it is an isoceles triangle. by using st1 then why this is not sufficient.

Thanks

Perpendicular to hypotenuse is a perpendicular from A to BD.

As for why the first statement is not sufficient. All we know that one of the legs is 5, nothing else. HOW are you going to find BD? BD could take ANY length.
_________________

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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31 Jul 2014, 11:58

Bunuel wrote:

chiragatara wrote:

If angle BAD is a right angle, what is the length of side BD?

(1) AC is perpendicular to BD

(2) BC = CD

Official answer explanation:

Using statements 1 and 2, we know that AC is the perpendicular bisector of BD. This means that triangle BAD is an isosceles triangle so side AB must have a length of 5 (the same length as side AD). We also know that angle BAD is a right angle, so side BD is the hypotenuse of right isosceles triangle BAD. If each leg of the triangle is 5, the hypotenuse (using the Pythagorean theorem) must be 5 underroot 2.

Can someone kindly explainthe underlined portion?

For such kind of graphic questions you MUST post the image. Next, please also do check the OA's when posting a question, OA for this one is C, not E.

Original question is below:

If angle BAD is a right angle, what is the length of side BD?

Attachment:

triangleABCD.jpg

(1) AC is perpendicular to BD (2) BC = CD

Now, obviously each statement alone is not sufficient.

When taken together we have that AC is a perpendicular bisector. Now, if a line from a vertex to the opposite side is both perpendicular to it and bisects it then this side is a base of an isosceles triangle (or in other words if a bisector and perpendicular coincide then we have an isosceles triangle). You can check this yourself: in triangles ACD and ACB two sides are equal (AC=AC and BC=CD) and included angle between these sides are also equal (<ACD=<ACB=90) so we have Side-Angle-Side case (SAS), which means that ACD and ACB are congruent triangles, so AB=AD --> ABD is an isosceles triangle.

Next, as ABD is an isosceles triangle then AB=AD=5 and hypotenuse \(BD=5\sqrt{2}\).

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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04 Aug 2015, 22:07

As per option B, C is the midpoint of BD, and any line drawn from a vertex to midpoint of another side is median. So, Ac is median. I want to know that won't the property:- "Median drawn to the hypotenuse of right triangle is half length of hypotenuse length", applicable here? Please let me know where i am wrong? According to me the answer shud be B.

As per option B, C is the midpoint of BD, and any line drawn from a vertex to midpoint of another side is median. So, Ac is median. I want to know that won't the property:- "Median drawn to the hypotenuse of right triangle is half length of hypotenuse length", applicable here? Please let me know where i am wrong? According to me the answer shud be B.

The property you quote is right but how does this help to find the length of BD?
_________________

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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09 Sep 2015, 08:45

1

This post received KUDOS

Kekki wrote:

Isn't it that in an right angle triangle a perpenidcular line to hupothenuses (BC) halves the 90 degree triangle into 45° and 45° degrees?

No, not necesssarily. What you are mentioning is a particular case of the perpendicular drawn from A to BD also bisecting the side. When you do have the perpendicular drawn from the right angle to the opposite side also bisects the side, you have the triangle as 45,45,90.

If angle BAD is a right angle, what is the length of side [#permalink]

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17 May 2016, 08:36

Bunuel Can't we use the right angle property : Median of right angle is half the hypotenuse. Since statement 1 is already saying that AC is perpendicular it should be sufficient to say that the triangle is an isoceles triangle. Do we really need statement 2?

Bunuel Can't we use the right angle property : Median of right angle is half the hypotenuse. Since statement 1 is already saying that AC is perpendicular it should be sufficient to say that the triangle is an isoceles triangle. Do we really need statement 2?

How do you deduce that the triangle is isosceles? Please show your work.

As for why the first statement is not sufficient. All we know that one of the legs is 5, nothing else. HOW are you going to find BD? BD could take ANY length.
_________________

Re: If angle BAD is a right angle, what is the length of side [#permalink]

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17 May 2016, 09:10

Bunuel Thanks for your reply. Median from Angle BAD will divide the hypotenuse BD into two halves. And, from statement 1 we know that AC is perpendicular to BD so AC satisfies both conditions of being a perpendicular and a bisector. Therefore from these two conditions we know that Triangle BAD is isosceles triangle. Therefore, side AB is also 5 and it makes side BD as 5√2.