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.
Rectangle: a = w x l
Square: a = a^2
Triangle: a= b x h divided by 2
Parallelogram: a = HH
Trapezoid: a = a+b
2h
No, because
60/3 = 20
90/6= 15
Both of the answers aren’t equivalent.
