Geometry Problem 1525. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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To post a solution to this problem click Comment underneath the post, or click into the line that says, “Enter Comment.” Type what you want to say and press Publish to post your solution.

MN = 3

ReplyDeletehttps://photos.app.goo.gl/Qt4wWUDARG6fzTrbA

DeleteWhy the sketch is deleted

DeleteI'm sorry to hear that your sketch was deleted from the blog. there may have been a technical issue that led to the deletion. I have just re-sent your post to the blog. Thank you.

DeleteLet P be the mid-point of CD

ReplyDeleteIn Triangles BMN & CPN,

< MBN = < PCN (= A - 30)

BN = NC (since N is the circumcenter of right triangle ABC)

BM = MP (since Triangle DPM is equilateral)

So Triangles BMN & CPN are congruent SAS

Now using the midpoint theorem in Triangle ACD,

MN = NP = AD/2 = 6/2 = 3

Sumith Peiris

Moratuwa

Sri Lanka