The sides AB and BC of a square ABCD are produced to P and Q respectively so that BP= CQ. Prove that PD and AQ are equal and perpendicular to each other.
Answers
Step-by-step explanation:
Producing AB and BC to P and Q respectively, we get the above figure.
Given that BP=CQ
implies AB+BP=AB+CQ
we know that AB=BC (since ABCD is a square)
implies AB+BP=BC+CQ
implies AP=BQ. ->equation 1
we know that AD=AB. ->equation 2
squaring equation 1 and 2 on both sides and adding:
AP²+AD²=BQ²+AB²
applying Pythagoras theorem since DAP is a right angle and QBA is a right angle
PD²=AQ²
implies PD=AQ since distance can be only positive
hence, PD=AQ proved
we know that triangles APD and BQA are congruent to each other by SSS congruency. (considering equation 1 and 2 and PD=AQ)
Now, let us assume angle APD=x
This means angle BQA=x as the triangles APD and BQA are congruent.
we know that angle ABQ=90°
sum of angle in triangle ABQ=180°
QAB+ABQ+BQA=180°
QAB+90°+x=180°
QAB=90°-x°
Let us consider the point of intersection of AQ and DP as O
The sum of angles in triangle AOP=180°
AOP+OPA+PAO=180°
AOP+x+90°-x=180° (angle OPA=DPA and angle PAO=QAP=QAB)
AOP+90°=180°
AOP=90°
we can see that the angle between line segments PD and AQ is 90°
Hence, PD and AQ are perpendicular to each other
