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10. 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.

Hello Ajay, we have to take two triangles ABQ and ADP
<ABQ = <PAD = 90
Side AD = side AB
Also AB+BP = AP 
AB = BC (side of square)
BP = CQ ( given)
So AP = BC + CQ = BQ
By SAS rule both triangles are congruent
So by CPCT rule, side PD = AQ [PROVED]
In triangle, ARP, ext <PRQ = sum of interiors <RAP + <RPA
Now <RAP + <AQB = 90
Because of congruency <AQB = < RPA
So <RAP + <RPA = 90 = <PRQ
Hence AQ and PD are normal to each other [PROVED]

 

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In triangle AQB and ADP
AB common
BC=AB (SQUARE)
SO,BC+CQ=AB+BP
Angle DAB=Angle ABC
So triangle AQB =~ Congurent ADP
SO DP =AQ
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