Question

in a triangle abc bd is perpendicular to ac and ce perpendicular to ab if bd and ce intersect at o prove that angle boc is 180-a

Answer 1

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Answer 2

Answer:

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

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