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UPSC CDS 2 2021 Math Paper

Directions (Q.Nos. 54 and 55) Consider the following next two questions that follow.

A B C

is triangle in which

A B=A C

and

D

is any point on

B C

Question : 54 of 100

Marks: +1, -0

Which one of the following is correct?

Solution: 👈: Video Solution

Given,

A\;B\;C

is a triangle in which

A\;B=A\;C

and

D

is any point on

B\;C

Draw

A\;M\perp B\;C

\triangle ABM

and

\triangle ACM

A\;B=A\;C\;\;

(given)

\angle 1=\angle 2

(90^{\circ})

and

A\;M=A\;M

(common)

\Rightarrow \triangle ABM \cong \triangle ACM

\therefore B\;M=C\;M \;\; [ \because \text{CPCT} ] \;\; \dots

(i)

\;\text{ Now, }\;\; A\;B^{2}=A\;M^{2}+B\;M^{2}\;\dots\;\text{ (ii) }\;

\;\; \;\text{ Also, }\; A\;D^{2}= \;\; A\;M^{2}+D\;M^{2} \;\; \dots \;\text{ (iii) }\;

A\;B^{2}-A\;D^{2}= \;\; (A\;M^{2}+B\;M^{2})

\;\; -(A\;M^{2}+D\;M^{2})

= \;\; A\;M^{2}+B\;M^{2}-A\;M^{2}-D\;M^{2}

= \;\; B\;M^{2}-D\;M^{2}=(B\;M+D\;M)(B\;M-D\;M)

= \;\; (C\;M+D\;M)(B\;M-D\;M) \;\; \text{ (from Eq. (i)) }\;

= \;\; C\;D \times B\;D

\;\text{ Hence, }\; A\;B^{2}-\; A\;D^{2}=C\;D \times B\;D \;\text{ Or }\;

A\;C^{2}-A\;D^{2}= \;\; C\;D \times B\;D

\; \;\; [\;\text{ since, }\; A\;B=A\;C]

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