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Standard IX

Mathematics

Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

In the given ...

Question

In the given fig. $A B C D$ is a square, $M$ is the midpoint of $A B$ and $P Q ⊥ C M$ meets $A D$ at $P$ and $C B$ produced to $Q$ . Prove that:
(a) $P A = B Q$
(b) $C P = A B + P A$

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Solution

$\begin{matrix} Δ P A M a n d Δ B M Q ∠ P M A = ∠ B M Q = α A M = M B ∠ P A M = ∠ M B Q = 90^{\circ} I n Δ C P Q L M ⊥ P Q P M = M Q C P = C Q (i s o s c e l e s Δ C P Q) C Q = C B + B Q C Q = A B + B Q S o, B A = B Q \end{matrix}$

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In the given fig. ABCD is a square, M is the midpoint of AB and PQ⊥CM meets AD at P and CB produced to Q. Prove that:(a) PA=BQ(b) CP=AB+PA

ΔPAMandΔBMQ∠PMA=∠BMQ=αAM=MB∠PAM=∠MBQ=90∘InΔCPQLM⊥PQPM=MQCP=CQ(isoscelesΔCPQ)CQ=CB+BQCQ=AB+BQSo,BA=BQ

In the given fig. $A B C D$ is a square, $M$ is the midpoint of $A B$ and $P Q ⊥ C M$ meets $A D$ at $P$ and $C B$ produced to $Q$ . Prove that:
(a) $P A = B Q$
(b) $C P = A B + P A$

$\begin{matrix} Δ P A M a n d Δ B M Q ∠ P M A = ∠ B M Q = α A M = M B ∠ P A M = ∠ M B Q = 90^{\circ} I n Δ C P Q L M ⊥ P Q P M = M Q C P = C Q (i s o s c e l e s Δ C P Q) C Q = C B + B Q C Q = A B + B Q S o, B A = B Q \end{matrix}$