Problem:
In triangle ABC,∠ABC=120∘,AB=3 and BC=4. If perpendiculars constructed to AB at A and to BC at C meet at D, then CD=
Answer Choices:
A. 3
B. 3​8​
C. 5
D. 211​
E. 3​10​
Solution:
Extend CB and DA to meet at E. Since ∠E=30∘, EB=6. Hence EC=10 and CD=3​10​. In general, if AB=x and BC=y then EB=2x and EC=2x+y, so CD=3​2x+y​.
OR
Draw a line through B parallel to AD intersecting CD at H. Then drop a perpendicular from H to I on AD. Note that BHC and HDI are both 30∘−60∘−90∘ triangles. Thus CH=3​4​. Since HI=AB=3, it follows that HD=3​6​. Hence CD=CH+HD=3​4​+3​6​=3​10​.
OR
Draw a line through C parallel to AD and meeting AB extended at F. Then ∠BCF=30∘, so BF=2. Drop a perpendicular from C to G on AD. Since AFCG is a rectangle, CG=AF=AB+BF=5. Since ∠CDG=60∘, we have CD=3​2​CG=3​10​.
Note. One can also draw AC, apply the Law of Cosines to both â–³ABC and â–³ADC, and then equate the resulting values of AC to find CD.
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