GMAT Club Forumhttps://gmatclub.com:443/forum/ In the figure, BD and CD are angular bisectors of

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

 Attachments: triangle.png [ 6.55 KiB | Viewed 2947 times ]

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

 Attachments: triangle.png [ 2.29 KiB | Viewed 2853 times ]

 Author: ThiruN [ 26 May 2018, 20:21 ] Post subject: Re: In the figure, BD and CD are angular bisectors of

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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