Q.3) ABCD is a parallelogram. P and Q are point
on the diagonal AC such that DP and Bo are
perpendiculars to AC.
Prove that DP=BO
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Step-by-step explanation:
consider triangle APD and triangle BQC
AD=BC ( because in parallelogram two opposites sides are equal)
angle DPA = angle BQC ( both are 90°)
angle DAP = angle BCQ ( as they are alternate angles)
therefore by ASA theorem triangle APD and triangle BQC are similar
hence DP = BO
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