A triangle has corners at points A, B, and C. Side AB has a length of #21 #. The distance between the intersection of point A's angle bisector with side BC and point B is #5 #. If side AC has a length of #16 #, what is the length of side BC?

Question

A triangle has corners at points A, B, and C.  Side AB has a length of #21  #.  The distance between the intersection of point A's angle bisector with side BC and point B is #5  #.  If side AC has a length of #16  #, what is the length of side BC?

Maverick Smith · Accepted Answer

BC ≈ 8.81

Firstly, let the point where the angle bisector intersects with side BC be D. Then by the #color(blue)" Angle bisector theorem " # # (BD)/(DC) = (AB)/(AC) # Require to find DC. Substitute the appropriate values into the ratio to obtain. #rArr 5/(DC) = 21/16 # Now cross-multiply : #21xxDC = 16xx5 # To obtain DC , divide both sides by 21 #rArr (cancel(21) DC)/cancel(21) = (16xx5)/21 # # rArr DC ≈ 3.81# Now , BC = BD + DC = 5 + 3.81 = 8.81 to 2 decimal places.

Ella Armstrong · Answer

undefined Using the angle bisector theorem, the ratio of the lengths of the segments of side AC divided by the length of side AB equals the ratio of the lengths of the segments of side BC divided by the length of side BC.

Thus, $ \frac{AC}{AB} = \frac{BC}{BC} $.

Solving for BC, we get $ BC = \frac{AB \cdot BC}{AC} $.

Given that AB = 21 and AC = 16, plugging in these values yields $ BC = \frac{21 \cdot BC}{16} $.

Multiplying both sides by 16, we get $ 16 \cdot BC = 21 \cdot BC $.

Dividing both sides by BC, we get $ 16 = 21 $.

Therefore, BC = 16.