Question

in a triangle abc , bd and ce are perpendicular to the sides ac and ab respectively. if bd and ce intersect at i prove that: ∆bic =180-∆a

Answer 1

In triangle ABC,□A+□B+□C=180°(angles sum property)

                          □B+□C=180°-□A ...........(1)

 In triangle OBC,□BOC+□OBC+□OCB=180°(angles sum property)

                             □OBC+□OCB =180°-□BOC ..........(2)

Now, In triangle BEC

                                □BEC+□CBE+□BCE=180°

                                  90°+□CBE+□BCE=180°

                                  □CBE+□BCE=90°............(3)

Similarly,In triangle BCD

                                  □BCD+□DBC =90°............(4)

Adding (3) and  (4),we get

□CBE+□BCE+□BCD+□DBC=180°

□B+□OCB+□C+□OBC=180°

□B+□C+□OCB+□OBC=180°

180°-□A+180°-□BOC=180° (from 1 and 2 )

□A=180°-□BOC ........Proved