AB is a line segment-P and Q are points on oposite sides of AB such that each of them is equidistant from the points A and B-Show that triangle-PAQ-congtriangle-PBQ-

Question

Toppr Admin · Accepted Answer

In -x25B3-PAQ and -x25B3-PBQPA-PB-given-QA-QB-given-PQ-PQ-common-So-x25B3-PAQ-x2245-x25B3-PBQ by SSS rule-

$A B$ is a line segment. $P$ and $Q$ are points on oposite sides of $A B$ such that each of them is equidistant from the points $A$ and $B$ .Show that $△ P A Q ≅ △ P B Q$

In $△ P A Q$ and $△ P B Q$

$P A = P B$ (given)

$Q A = Q B$ (given)

$P Q = P Q$ (common)

So, $△ P A Q ≅ △ P B Q$ by SSS rule.

AB is a line segment.P and Q are points on oposite sides of AB such that each of them is equidistant from the points A and B.Show that △PAQ≅△PBQ

In △PAQ and △PBQPA=PB(given)QA=QB(given)PQ=PQ(common)So,△PAQ≅△PBQ by SSS rule.

$A B$ is a line segment. $P$ and $Q$ are points on oposite sides of $A B$ such that each of them is equidistant from the points $A$ and $B$ .Show that $△ P A Q ≅ △ P B Q$

In $△ P A Q$ and $△ P B Q$

$P A = P B$ (given)

$Q A = Q B$ (given)

$P Q = P Q$ (common)

So, $△ P A Q ≅ △ P B Q$ by SSS rule.