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| In the figure, BD and CD are angular bisectors of <B and <C in an https://gmatclub.com/forum/in-the-figure-bd-and-cd-are-angular-bisectors-of-b-and-c-in-an-187432.html |
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| Author: | vatsavayi [ 25 Oct 2014, 05:12 ] | ||
| Post subject: | In the figure, BD and CD are angular bisectors of <B and <C in an | ||
In the figure, BD and CD are angular bisectors of <B and <C in an equilateral triangle ABC. What is the ratio of the area polygon ABDC to the area of triangle BDC? A) 1:2 B) 2:3 C) 3:2 D) 2:1 E) 3:1
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| Author: | PareshGmat [ 27 Oct 2014, 20:43 ] | ||
| Post subject: | Re: In the figure, BD and CD are angular bisectors of <B and <C in an | ||
Join point A to point D \(\triangle\) ABD = \(\triangle\) ACD = \(\triangle\) BDC Area polygon ABDC = 2 times Area \(\triangle\) BDC Ratio = 2:1 Answer = D
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| Author: | ThiruN [ 26 May 2018, 20:21 ] |
| Post subject: | Re: In the figure, BD and CD are angular bisectors of <B and <C in an |
How do we come to the conclusion that △ ABD = △ ACD = △ BDC? |
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| Author: | kaiserkaran [ 03 Jun 2018, 12:17 ] | ||
| Post subject: | In the figure, BD and CD are angular bisectors of <B and <C in an | ||
ThiruN wrote: How do we come to the conclusion that △ ABD = △ ACD = △ BDC? hi, we know that BD and CD are angular bisectors and from the figure we know they meet at point D, which is basically the incenter. If an angle bisector is drawn from A it meets at D as show in the previous post. if u find the angles for all the points of both the triangles u will find that the bigger triangle gets broken down to 3 parts which are equal to each other.
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