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# In the figure above, D is a point on the side AC of ABC.Is ABC isos?

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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
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In the figure above, D is a point on side AC of ΔABC. Is ΔABC is isosceles?

(1) The area of triangular region ABD is equal to the area of triangular region DBC --> this gives us info that D is the midpoint of AC but we don't have any info about angle ADB & BDC....insufficient
(2) BD┴AC and AD = DC --> this gives us info that D is midpoint of AC and that BD┴AC, this condition will make AB=BC.......sufficient.(refer below image, hope my understanding is correct )

Ans. B)
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
What exactly does this "BD┴AC" say?

I don't know what the symbol means.
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
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noTh1ng wrote:
What exactly does this "BD┴AC" say?

I don't know what the symbol means.

It means that BD is perpendicular to AC.
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
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Nevernevergiveup wrote:
Attachment:
1.png

In the figure above, D is a point on the side AC of ΔABC. Is ΔABC isosceles?

(1) The area of the triangular region ABD is equal to the area of triangular region DBC.

(2) BD is perpendicular to AC and AD = DC.

Statement 1 is straightforward to see that it is NOT sufficient. The most this statement will do is to provide a scenario when we have AD=CD. But this does not yet answer the question asked.

Statement 2, is tricky as in from the figure provided, BD can either be to the left of A or between A and C or to the right of C. But in only 1 case (of between A and C) will you get AD=CD such that BD $$\perp$$ AC. With this scenario, triangle ABC is indeed isosceles (by making triangles ABD and BCD congruent to each other by the SAS property ---> AB=BC) . Hence sufficient.

Hope this helps.
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]

We are after: Is ΔABC isosceles?

(1) Not sufficient. Using the formula for areas of triangle, we get

Area of ΔABD = Area of ΔABD
1/2 * AD * h = 1/2 * DC * h

h is the height for both triangles, if you drop a perpendicular from point B on AC. This will fall on the extended AC (if you can imagine it)

The above equation yields to:

But we want to know if ΔABC is isosceles. This information doesn't help so it is INSUFFICIENT.

(2) Sufficient

Since BD ⊥AC, and D is on AC, imagine D dividing AC into two sides i.e. AD =AC.

Now imagine two right angled Δs inside ABC being split in the middle by BD. These two right angled Δs have the same base and height so the hypotenuse will be same i.e. AB = AC
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
Bunuel VeritasKarishma chetan2u Cant option B result in equilateral triangle?
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
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Bishal123456789 wrote:
Bunuel VeritasKarishma chetan2u Cant option B result in equilateral triangle?

Hi Bishal,

Yes, it CAN be an equilateral triangle....
Of course, it is an isosceles triangle for sure.. also all equilateral triangles are Isosceles
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
A strong takeaway in this Q is that ALL equilateral triangles are ALSO isosceles triangle. If we forget this rule, we might reject option B.

The triangle property is ' When an perpendicular bisects the base into equal parts (AD=DC for example as per the Q), then such triangle (ABC) must be an isosceles or equilateral triangle.
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
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Re: In the figure above, D is a point on the side AC of ABC.Is ABC isos? [#permalink]
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