- Thread starter Qwerty21
- Start date

- Joined
- Mar 16, 2016

- Messages
- 2,043

BC is a common side

Thank you, do you have another answer?BC is a common side

- Joined
- Jun 18, 2007

- Messages
- 22,405

Please tell us how you used the information provided in response #2.Thank you, do you have another answer?

Are you familiar with theorem for "angle bisector of a triangle"?

The supplementary, complementary? Like that? (sorry if my grammar is wrong)Please tell us how you used the information provided in response #2.

Are you familiar with theorem for "angle bisector of a triangle"?

- Joined
- Jun 18, 2007

- Messages
- 22,405

I cannot understand your response.The supplementary, complementary? Like that? (sorry if my grammar is wrong)

Are you familiar with theorem for "angle bisector of a triangle"?

If you are - please state it.

I searched it..I cannot understand your response.

Are you familiar with theorem for "angle bisector of a triangle"?

If you are - please state it.

In geometry, the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

- Joined
- Mar 16, 2016

- Messages
- 2,043

- Joined
- Jun 18, 2007

- Messages
- 22,405

I searched it..

In geometry, the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

In your problem, the line BC is the bisector and it divides base AD into AB and BC. What do you deduce about ratio AB:BD?

You are given AC=CD. What does that lead to about AB:BD ?

- Joined
- Nov 12, 2017

- Messages
- 9,133

I don't think this is a vector problem. Those are line segments, not vectors; and they are congruent, not equal.The diagram doesn't seem to match with the vector expression \(\displaystyle \overline{AC}\equiv\overline{DC} \) which implies that point A and D are exactly the same point.

- Joined
- Nov 12, 2017

- Messages
- 9,133

I think you are telling us that you have not learned this theorem, but had to search for it. Therefore, it is not what you are expected to be using.I searched it..

In geometry, the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

This suggests that perhaps you observed that angles ACB and DCB are supplementary. That's a reasonable thing to think about, considering that the information you are given is SSA, and those angles would be interesting to work with. (It's possible that your work might replicate part of the proof of the angle bisector theorem!)The supplementary, complementary? Like that? (sorry if my grammar is wrong)

But the problem evidently is intended to use the ASA postulate. (Is that true?) Would you like to proceed using that, rather than the angle bisector theorem? If so, it will be very helpful if you show us what you have tried.