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  • In the given figure, ABCD is a square, M is the midpoint of A

In the given figure,

ABCD is a square, M is the midpoint of AB and \ P Q \perp C M meets AD at P and BC produced at Q. Then,

Option: 1

C P<A B+P A


Option: 2

C P=A B+P A


Option: 3

C P>A B+P A


Option: 4

None of the above


Answers (1)

best_answer

\begin{align*} &\text{In } \triangle PAM \text{ and } \triangle QBM, \text{ we have:} \\ &AM = BM \quad (\because M \text{ is the midpoint of } AB) \\ &\angle PAM = \angle QBM \quad \left(\text{each equal to } 90^{\circ}\right) \\ &\angle AMP = \angle BMQ \quad \text{(vertically opp. } \angle\text{s)} \\ \therefore &\triangle PAM \cong \triangle QBM \quad \text{(by AAS-criteria)} \\ \text{Hence, } &PM = QM \quad \text{(c.p.c.t.)} \end{align*}

\\\text{Now, in } \triangle C M P\text{ and }\Delta C M Q,\text{ we have} \\\\\ P M=Q M \quad(\because \triangle P A M \cong \Delta Q B M) \\\\ \angle C M P=\angle C M Q \quad\text{ (each equal to }90^{\circ} ) \\\\ C M=C M \quad\text{( common )} \\\\ \therefore \Delta C M P \cong \Delta C M Q \qquad\text { (SAS-criteria). } \\\\ \therefore C P=C Q \qquad \text { (c.p.c.t.). } \\\\ \Rightarrow C P=C B+Q B=A B+P A \quad[\because C B=A B \text { and } Q B=P A] \\\\ \text {Hence, } C P=A B+P A \text { . }

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Ritika Jonwal

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