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Question

In a right angled ABC, right angled at A, if ADBC such that AD=p, if BC=a,CA=b and AB=c, then:
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A
p2=b2+c2
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B
1p2=1b2+1c2
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C
pa=pb
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D
p2=b2c2
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Solution

The correct option is B 1p2=1b2+1c2
Given, In ABC, A=90 and ADBC
In ABC,

BAC+ABC+ACB=180

90+ABC+ACB=180

ABC+ACB=90 (I)
In CAD,

CAD+ACD+ADC=180

CAD+ACD+90=180

CAD+ACD=90..(II)

Equating (I) and (II),
ABC+ACB=CAD+ACD
ABC=CAD...(III)
Similarly, ACB=BAD...(IV)
Now, In s, ABC and DAC

ABC=CAD ..(From III)

BAC=CDA (Each 90)

ACD=ACB (Common angle)

Thus, ABCDAC (AAA rule)

Thus, ACDC=BCAC (Sides of similar triangles are in proportion)

AC2=BC×DC... (V)
Similarly, ABCDBA

and ABBD=BCAB (Sides of similar triangles are in proportion)

AB2=BD×BC ....(VI)
Similarly, ABDCAD (AAA rule)

Thus, ADCD=BDAD (Sides of similar triangles are in proportion)

AD2=BD×CD
Now,
1AB2+1AC2=1BC×DC+1BD×BC

1AB2+1AC2=1BC(1DC+1BD)

1AB2+1AC2=1BC(BD+DCBD×DC)

1AB2+1AC2=1BC(BCBD×DC)

1AB2+1AC2=1BD×DC

1AB2+1AC2=1AD2

1c2+1b2=1p2

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