Question

In ΔABC, point N is on AB and point M is on AC such that, A-N-B, A-M-C and BN = CM. Line NM intersects line BC in L and B-C-L. Prove that, $\frac{LM}{LN}$] = $\frac{AB}{AC}$.

Answer 1

Answer:

bee gees is your answer and follow the above mentioned

Answer 2

Step-by-step explanation:

AB = AC (Sides opposite to equal angle are equal) Subtracting BM from both sides, we get AB – BM = AC – BM ⟹AB – BM = AC – CN (∵BM =CN) ⟹AM =AN ∴∠AMN =∠ ANM (Angles opposite to equal sides are equal) Now, in ∆ABC, ∠A+ ∠B + ∠C =1800 ----(1) (Angle Sum Property of triangle) Again In ∆AMN, ∠A + ∠AMN + ∠ ANM =1800 ----(2) (Angle Sum Property of triangle) From (1) and (2), we get ∠B + ∠C = ∠ AMN + ∠ ANM ⟹ 2∠B = 2∠ AMN ⟹∠B = ∠ AMN Since, ∠B and ∠ AMN are corresponding angles. Read more on Sarthaks.com - https://www.sarthaks.com/134647/in-abc-and-are-points-the-sides-and-ac-respectively-such-that-bm-cn-if-b-c-then-show-that-mn?show=134660#a134660