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>>Some more Congruence Criteria

>>In the given figure, ABCD is a square an

Question



In the given figure, ABCD is a square and P is a point inside it such that PB=PD. Prove that CPA is a straight line.





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It is given that ABCD is a square and P is a point inside it such that PB=PD

Considering △APD and △APB

We know that all the sides are equal in a square

So we get DA=AB

AP is common i.e. AP=AP

According to SSS congruence criterion

△APD≅△APB

We get ∠APD=∠APB(c.p.c.t)…(1)

Considering △CPD and △CPB

We know that all the sides are equal in a square

So we get CD=CB

CP is common i.e. CP=CP

According to SSS congruence criterion

△CPD≅△CPB

We get ∠CPD=∠CPB(c.p.c.t)…(2)

By adding both the equation (1) and (2)

∠APD+∠CPD=∠APB+∠CPB…(3)

From the figure we know that the angles surrounding the point P is 360∘

So we get

∠APD+∠CPD+∠APB+∠CPB=360∘

By grouping we get

∠APB+∠CPB=360∘(∠APD+∠CPD)…(4)

Now by substitution of (4) in (3)

∠APD+