This Blog exists for the collective benefit of all geometry students. All questions are welcome. The more specific your question (including your own attempts to answer it) the better.
EVEN MORE WELCOME ARE ANSWERS FROM FELLOW STUDENTS. BLOG ON!
The challenge is to "see" the similar triangles... clearly you see the right angles... can you find the triangles that include the right angles and share angle-A?
Trngl-EHA is similar to Trngl-GDA based on AA, yes? From that you can build a proportion that will lead you to the proof using the means-extremes property.
dont get #27 on hw
ReplyDeleteCHRIS
or 29
ReplyDeleteCHRIS
Hey Chris,
ReplyDeleteAs for #27...
The challenge is to "see" the similar triangles... clearly you see the right angles... can you find the triangles that include the right angles and share angle-A?
Trngl-EHA is similar to Trngl-GDA based on AA, yes? From that you can build a proportion that will lead you to the proof using the means-extremes property.
As for #29...
ReplyDeleteSimilar to #27, can you can find two triangles that include the congruent angles (1&2) and that share angle-A?
Trngl-CAB is similar to trngl-BAD based on AA. Once again, you should be able to build a proportion that will support the proof.
By definition, you know that corresponding sides of similar triangles are in proportion.