The given figure shows a - ABC in which AB-AC and BP-CQ Then- - ABQ - ACP

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Standard VII

Mathematics

Equal Angles Subtend Equal Sides

The given fig...

Question

The given figure shows a $Δ A B C$ in which $A B = A C$ and $B P = C Q$ Then,
$Δ A B Q ≅ Δ A C P$

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Solution

Given, $B P = C Q$

$B P + P Q = C Q + P Q$

$B Q = C P$ ..(I)

In $△$ ABQ and ACP

$A B = A C$ (Given)

$B Q = C P$ (From I)

$∠ A B Q = ∠ A C P$ (Isosceles triangle property)

Hence, $△ A B Q ≅ △ A C P$ (SAS postulate)

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The given figure shows a ΔABC in which AB=AC and BP=CQ Then,ΔABQ≅ΔACP

Given, BP=CQBP+PQ=CQ+PQBQ=CP..(I)In △ABQ and ACPAB=AC (Given)BQ=CP (From I)∠ABQ=∠ACP (Isosceles triangle property)Hence, △ABQ≅△ACP (SAS postulate)

The given figure shows a $Δ A B C$ in which $A B = A C$ and $B P = C Q$ Then,
$Δ A B Q ≅ Δ A C P$

Given, $B P = C Q$

$B P + P Q = C Q + P Q$

$B Q = C P$ ..(I)

In $△$ ABQ and ACP

$A B = A C$ (Given)

$B Q = C P$ (From I)

$∠ A B Q = ∠ A C P$ (Isosceles triangle property)

Hence, $△ A B Q ≅ △ A C P$ (SAS postulate)