In triangle $A B C$ ; $A B = A C . P, Q$ and $R$ are mid-points of sides $A B, A C$ and $B C$ respectively. Prove that:
$B Q = C P$

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Solution

$A B = A C$ [Given]
$\Rightarrow ∠ B = ∠ C$
Also,
$\frac{1}{2} A B = \frac{1}{2} A C$
$\Rightarrow B P = C Q$ [ $P$ and $Q$ are mid- points of $A B$ $A C$ ]
In $△ B P C$ and $△ C Q B,$
$B P = C Q q$
$∠ B = ∠ C$
$B C = B C$
Hence, $△ B P C ≅ △ C Q B$ [SAS]
$B Q = C P$ [C.P.C.T]

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In triangle ABC ; AB=AC. P,Q and R are mid-points of sides AB,AC and BC respectively. Prove that:BQ=CP

AB=AC [Given]⇒∠B=∠CAlso,12AB=12AC⇒BP=CQ [P and Q are mid- points of AB AC]In △BPC and △CQB,BP=CQq∠B=∠CBC=BCHence, △BPC≅△CQB [SAS]BQ=CP [C.P.C.T]

In triangle $A B C$ ; $A B = A C . P, Q$ and $R$ are mid-points of sides $A B, A C$ and $B C$ respectively. Prove that:
$B Q = C P$

$A B = A C$ [Given]
$\Rightarrow ∠ B = ∠ C$
Also,
$\frac{1}{2} A B = \frac{1}{2} A C$
$\Rightarrow B P = C Q$ [ $P$ and $Q$ are mid- points of $A B$ $A C$ ]
In $△ B P C$ and $△ C Q B,$
$B P = C Q q$
$∠ B = ∠ C$
$B C = B C$
Hence, $△ B P C ≅ △ C Q B$ [SAS]
$B Q = C P$ [C.P.C.T]