Two angles are said to be complementary, if the sum of the measures of the to angles is equal to 90 degrees.
Thus, given that HFG is complementary to ACB, them mHFG + mACB = 90 degrees.
From the figure, given that the line from point F meats line CE at point P, then HFG = CFP.
But mCFE = 90 degrees and mCFE = mCFP + mPFE
Also PFE = DFH
Thus, mCFE = mCFP + mPFE = mHFG + mDFH = 90 degrees
Recall that mHFG + mACB = 90 degrees
Thus, mHFG + mACB = mHFG + mDFH
Therefore, mACB = mDFH.
Answer:
Hello your question is poorly written attached below is the complete question
answer : attached below
Step-by-step explanation:
To Prove: Z is located 2/3 of the distance from each vertex of ΔABC to the midpoint of the opposite side. we will apply ; property of bisecting a line , equality theorem , transitive property and similarity theorem
Attached below is the proof
2.96 or 18.75 is the answer im leaning towards 2.96 though
Try this suggested option (all the details are in the attachment), the correct orientation is marked with red and green colours.
P.S. The point C has coordinates: (3;1). If to traslate it 6 units right and 5 units down, then (3+6;1-5) ⇒ (9;-4). The same principle is for the others points A, B and D. Note, after translation point A is point E, B⇒F, C⇒G and D⇒H.
Ok bet I got you say less