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Question: In triangle $ABC$, $D$ lies on side $BC$ such that $\overline{AD}$ bisects $\angle BAC$ and $\angle ADB$ is obtuse.

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Which of the following could be true? (There might be more than one correct answer.)

(A) $AD<BC$ and $CD \geq AB$

(B) $AD \geq BC$ and $CD<AB$

(C) $AD<BC$ and $CD<AB$

(D) $AD \geq BC$ and $CD \geq AB$

Here's the observations I made:

  • Using the angle bisector theorem, if $AC=a$ and $AB=b$, then we can say that $CD=ax$ and $BD=bx$.

  • In $\triangle ABD$, because $AB>DB$, it follows that $0<x<1$.

  • Using this fact in $\triangle ACD$, it follows that $\\angle CAD=\angle DAB>180^{\circ}-\angle ADB$.

Returning back to the original question, my idea was to consider the shared sides between the pairs of triangles relevant to the lengths mentioned in the options. However, all the work I've done with this has resulted in a dead end.

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    $\begingroup$ I did not check in Detail. It seems that all 4 are not necessarily true. We can make triangles where each is true. We can make alternate triangles where each is not true. You may want to check the Question Wording. You should give the Question Source too. $\endgroup$
    – Prem
    Jul 15, 2023 at 20:27
  • $\begingroup$ Apologies, I edited the question to make it more clear. $\endgroup$
    – Indecisive
    Jul 17, 2023 at 16:18

1 Answer 1

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Since $\angle ADB$ is obtuse, and $AD$ is the angle bisector, then $\angle ABC<\angle ACB$, so that $AC<AB$, and by the angle bisector theorem$$CD<DB$$And since in a triangle the greater side is opposite the greater angle, then in $\triangle ADB$,$$DB<AB$$Therefore$$CD<AB$$ Hence (B) and (C) are true, (A) and (D) are false: regardless of the size relationship of $AD$ and $BC$,$$CD<AB$$ angle bisector/obtuse angle ADB

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