in figure ABCD is a square. if angle PQR=90 and PB=QC=DR, prove that angle QPR=45
basnetbhavna:
ABCD I'd a figure then from where comes
from where does PQR comes. where is the figure ..........
the figure is available at: http://hi.static.z-dn.net/files/db5/4f6ba45ada8b65864fea7b2dc8e30eab.png
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According to figure at http://hi.static.z-dn.net/files/db5/4f6ba45ada8b65864fea7b2dc8e30eab.png
In the figure, if PB=QC=DR, then we can deduce that DR = QR. This means PQ = QR and the triangle PQR is right as there is a 90degree angle. Since the two sides are equal, they are isosceles. So the angles PQR and QRP must be the same. Since three angles' sum is 180, the angles are of measure (180 - 90) / 2 = 45degrees.
Hence proved.
In the figure, if PB=QC=DR, then we can deduce that DR = QR. This means PQ = QR and the triangle PQR is right as there is a 90degree angle. Since the two sides are equal, they are isosceles. So the angles PQR and QRP must be the same. Since three angles' sum is 180, the angles are of measure (180 - 90) / 2 = 45degrees.
Hence proved.
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